{"id":4433,"date":"2023-02-10T07:38:13","date_gmt":"2023-02-10T07:38:13","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4433"},"modified":"2023-02-10T07:38:57","modified_gmt":"2023-02-10T07:38:57","slug":"ncert-solutions-class-9-maths-chapter-8-quadrilaterals-ex-8-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-8-quadrilaterals-ex-8-1\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 8 (Quadrilaterals)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 8 Quadrilaterals <\/strong>Exercise 8.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 8: Quadrilaterals<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-8-quadrilaterals-ex-8-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 8.2<\/a><\/li>\n<\/ul>\n<div>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 8.1<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.<br \/>\n\u2234 3x + 5x + 9x + 13x = 360\u00b0\u00a0 [Angle sum property of a quadrilateral]<br \/>\n\u21d2 30x = 360\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x =\u00a0<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"msubsup\"><span id=\"MathJax-Span-5\" class=\"texatom\"><span id=\"MathJax-Span-6\" class=\"mrow\"><span id=\"MathJax-Span-7\" class=\"mn\">360\u00b0\/<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-11\" class=\"mn\">30<\/span><\/span><\/span><\/span><\/span>\u00a0= 12\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 3x = 3 \u00d7 12\u00b0 = 36\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">5x = 5 \u00d7 12\u00b0 = 60\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">9x = 9 \u00d7 12\u00b0 = 108\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">13x = 13 \u00d712\u00b0 = 156\u00b0<br \/>\n\u21d2 The required angles of the quadrilateral are 36\u00b0, 60\u00b0, 108\u00b0 and 156\u00b0.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its\u00a0interior angles\u00a0is 90\u00b0.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4438\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"247\" height=\"148\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans2.png 933w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans2-300x180.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans2-768x461.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans2-480x288.png 480w\" sizes=\"auto, (max-width: 247px) 100vw, 247px\" \/><br \/>\nIn \u2206ABC and \u2206DCB,<br \/>\nAC = DB\u00a0 \u00a0 \u00a0[Given]<\/span><br \/>\n<span style=\"color: #000000;\">AB = DC\u00a0 \u00a0 \u00a0[Opposite sides of a parallelogram]<\/span><br \/>\n<span style=\"color: #000000;\">BC = CB\u00a0 \u00a0 \u00a0[Common]<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2206ABC \u2245 \u2206DCB [By SSS congruency]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220ABC = \u2220DCB [By C.P.C.T.]\u00a0 &#8212;&#8212;&#8212;&#8212;- (1)<\/span><br \/>\n<span style=\"color: #000000;\">Now, AB || DC and BC is a transversal. [ \u2235 ABCD is a parallelogram]<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ABC + \u2220DCB = 180\u00b0\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;- (2) [Co-interior angles]<\/span><br \/>\n<span style=\"color: #000000;\">From (1) and (2), we have<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC = \u2220DCB = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">i.e., ABCD is a parallelogram having an angle equal to 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 ABCD is a rectangle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4439\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"293\" height=\"175\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans3.png 1022w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans3-300x179.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans3-768x458.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans3-480x286.png 480w\" sizes=\"auto, (max-width: 293px) 100vw, 293px\" \/><br \/>\n\u2234 In \u2206AOB and \u2206AOD, we have<br \/>\nAO = AO [Common]<br \/>\nOB = OD [O is the mid-point of BD]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOB = \u2220AOD [Each 90]<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2206AQB \u2245 \u2206AOD [By,SAS congruency]<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AB = AD [By C.P.C.T.]\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- (1)<br \/>\nSimilarly, AB = BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;- (2)<br \/>\nBC = CD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;- (3)<\/span><br \/>\n<span style=\"color: #000000;\">CD = DA\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- (4)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 From (1), (2), (3) and (4), we have<\/span><br \/>\n<span style=\"color: #000000;\">AB = BC = CD = DA<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the quadrilateral ABCD is a rhombus.<\/span><br \/>\n<span style=\"color: #000000;\">Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Show that the diagonals of a square are equal and bisect each other at right angles.<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Let ABCD be a square such that its diagonals AC and BD intersect at O.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4440\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"136\" height=\"140\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans4.png 723w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans4-291x300.png 291w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans4-480x495.png 480w\" sizes=\"auto, (max-width: 136px) 100vw, 136px\" \/><br \/>\nThus, we have to prove AC = BD, OA = OC, OB = OD, and \u2220AOB = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Let ABCD be a square.<\/span><br \/>\n<span style=\"color: #000000;\">Let the diagonals AC and BD intersect each other at a point O.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ABC and \u0394DCB,<\/span><br \/>\n<span style=\"color: #000000;\">AB = DC (Sides of a\u00a0square\u00a0are equal to each other)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC = \u2220DCB (All\u00a0interior angles\u00a0are of 90<sup>o<\/sup>)<\/span><br \/>\n<span style=\"color: #000000;\">BC = CB (Common side)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394ABC \u2245 \u0394DCB (By\u00a0SAS congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AC = DB (By\u00a0CPCT)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000;\">Hence, the diagonals of a square are equal in length.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394AOB and \u0394COD,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOB = \u2220COD (Vertically opposite angles)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABO = \u2220CDO (Alternate interior angles)<\/span><br \/>\n<span style=\"color: #000000;\">AB = CD (Sides of a square are always equal)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394AOB \u2245 \u0394COD (By\u00a0AAS congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AO = CO and OB = OD (By CPCT)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212; (2)<\/span><br \/>\n<span style=\"color: #000000;\">Hence, the diagonals of a square bisect each other.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394AOB and \u0394COB,<\/span><br \/>\n<span style=\"color: #000000;\">As we had proved that diagonals\u00a0bisect\u00a0each other,<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AO = CO<\/span><br \/>\n<span style=\"color: #000000;\">AB = CB (Sides of a square are equal)<\/span><br \/>\n<span style=\"color: #000000;\">BO = BO (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394AOB \u2245 \u0394COB (By\u00a0SSS congruency)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220AOB = \u2220COB (By CPCT)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212; (3)<\/span><br \/>\n<span style=\"color: #000000;\">However, \u2220AOB + \u2220COB = 180\u00b0 (Linear pair)<\/span><br \/>\n<span style=\"color: #000000;\">2\u2220AOB = 180\u00b0 [From equation (3)]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOB = 90\u00b0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;- (4)<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0from equation (1), (2) and (4), we see that the diagonals of a square are equal and bisect each other at\u00a0right angles.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4440\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"162\" height=\"167\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans4.png 723w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans4-291x300.png 291w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans4-480x495.png 480w\" sizes=\"auto, (max-width: 162px) 100vw, 162px\" \/><br \/>\nIt is given that the diagonals of ABCD are equal and bisect each other at\u00a0right angles. <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AC = BD, OA = OC, OB = OD, and \u2220AOB = \u2220BOC = \u2220COD = \u2220AOD = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">To prove ABCD is a square, we have to prove that ABCD is a parallelogram in which AB = BC = CD = AD, and one of its interior angles is 90\u00b0. <\/span><br \/>\n<span style=\"color: #000000;\">In \u0394AOB and \u0394COD, <\/span><br \/>\n<span style=\"color: #000000;\">AO = CO (Diagonals bisect each other)<\/span><br \/>\n<span style=\"color: #000000;\">OB = OD (Diagonals bisect each other)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOB = \u2220COD (Vertically opposite angles)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394AOB \u2245 \u0394COD (SAS congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AB = CD (By CPCT)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (1)<\/span><br \/>\n<span style=\"color: #000000;\">And, \u2220OAB = \u2220OCD (By CPCT) <\/span><br \/>\n<span style=\"color: #000000;\">However, these are alternate interior angles for line AB and CD and\u00a0alternate interior angles are equal to each other only when the two lines are parallel. <\/span><br \/>\n<span style=\"color: #000000;\">Thus, AB || CD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (2)<\/span><br \/>\n<span style=\"color: #000000;\">From Equations (1) and (2), we obtain ABCD is a parallelogram. <\/span><br \/>\n<span style=\"color: #000000;\">In \u0394AOD and \u0394COD, <\/span><br \/>\n<span style=\"color: #000000;\">AO = CO (Diagonals bisect each other)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AOD = \u2220COD (Each angle is\u00a090\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">OD = OD (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394AOD \u2245 \u0394COD (SAS congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AD = DC (By CPCT)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8211; (3)<\/span><br \/>\n<span style=\"color: #000000;\">However, AD = BC and AB = CD (Opposite sides of\u00a0parallelogram\u00a0ABCD are equal)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AB = BC = CD = DA<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, all the sides of quadrilateral ABCD are equal to each other.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ADC and \u0394BCD,<\/span><br \/>\n<span style=\"color: #000000;\">AD = BC (Already proved)<\/span><br \/>\n<span style=\"color: #000000;\">AC = BD (Given)<\/span><br \/>\n<span style=\"color: #000000;\">DC = CD (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394ADC \u2245 \u0394BCD (SSS\u00a0Congruence rule) <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ADC = \u2220BCD (By CPCT) <\/span><br \/>\n<span style=\"color: #000000;\">However, \u2220ADC + \u2220BCD = 180\u00b0(Co-interior angles)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADC + \u2220ADC = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">2\u2220ADC = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220ADC = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">One of the interior angles of quadrilateral ABCD is a right angle.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, we have obtained that ABCD is a parallelogram where\u00a0AB = BC = CD = AD and one of its interior angles is 90\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, ABCD is a square.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Diagonal AC of a parallelogram ABCD bisects \u2220A (see figure). Show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4441\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"156\" height=\"152\" \/><\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i)<\/strong> it bisects \u2220C also,<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii)<\/strong> ABCD is a rhombus.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Given: The diagonal AC of a\u00a0parallelogram\u00a0ABCD bisects \u2220A.<br \/>\n<strong>(i) it bisects \u2220C also<br \/>\n<\/strong>ABCD is a parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DAC = \u2220BCA (Alternate\u00a0interior angles)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8211; (1)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAC = \u2220DCA (Alternate interior angles)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8211; (2)<\/span><br \/>\n<span style=\"color: #000000;\">However, it is given that AC bisects \u2220A.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DAC = \u2220BAC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (3)<\/span><br \/>\n<span style=\"color: #000000;\">From equations (1), (2), and (3), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DCA\u00a0 = \u2220BAC = \u2220DAC = \u2220BCA\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8211; (4)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220DCA = \u2220BCA<\/span><br \/>\n<span style=\"color: #000000;\">Hence, AC bisects \u2220C.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) ABCD is a rhombus.<br \/>\n<\/strong>From Equation (4), we obtain<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DAC = \u2220DCA<\/span><br \/>\n<span style=\"color: #000000;\">DA = DC (Side opposite to equal angles are equal)<\/span><br \/>\n<span style=\"color: #000000;\">However, DA = BC and AB = CD (Opposite sides of a parallelogram\u00a0 are equal)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, AB = BC = CD = DA<\/span><br \/>\n<span style=\"color: #000000;\">Hence, ABCD is a\u00a0rhombus.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. ABCD is a rhombus. Show that diagonal AC bisects \u2220Aas well as \u2220C and diagonal BD bisects \u2220B as well AS \u2220D.<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong><strong>Given:<\/strong> ABCD is a rhombus.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4442\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q7.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"273\" height=\"150\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q7.png 1022w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q7-300x165.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q7-768x422.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q7-480x264.png 480w\" sizes=\"auto, (max-width: 273px) 100vw, 273px\" \/><br \/>\nLet us join AC.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ABC, BC = AB (Sides of a\u00a0rhombus\u00a0are equal to each other)<\/span><br \/>\n<span style=\"color: #000000;\">So, \u2220BAC = \u2220BCA (Angles opposite to equal sides of a triangle are equal)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8211;(1)<\/span><br \/>\n<span style=\"color: #000000;\">However, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAC = \u2220DCA (Alternate interior angles for parallel lines AB and CD)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;-(2)<\/span><br \/>\n<span style=\"color: #000000;\">From equation (1) and (2), \u2220BCA = \u2220DCA <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AC\u00a0bisects\u00a0\u2220C.<\/span><br \/>\n<span style=\"color: #000000;\">Also, \u2220BCA = \u2220DAC (Alternate interior angles for parallel\u00a0lines BC and DA)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220BAC = \u2220DAC<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, AC bisects \u2220A as well as \u2220C.<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, it can be proved that BD bisects \u2220B and \u2220D as well.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. ABCD is a rectangle in which diagonal AC bisects \u2220A as well as \u2220C. Show that<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i)<\/strong> ABCD is a square<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii)<\/strong> diagonal BD bisects \u2220B as well as \u2220D.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>We have a rectangle ABCD such that AC bisects \u2220A as well as \u2220C.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4443\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"194\" height=\"116\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q8.png 1022w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q8-300x179.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q8-768x458.png 768w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q8-480x286.png 480w\" sizes=\"auto, (max-width: 194px) 100vw, 194px\" \/><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i) ABCD is a square<br \/>\n<\/strong>We are given that ABCD is a rectangle, so<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A = \u2220C<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220A = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u2220C<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220DAC = \u2220DCA (Given that AC\u00a0bisects\u00a0\u2220A and \u2220C)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, CD = DA (Sides opposite to equal angles are also equal)<\/span><br \/>\n<span style=\"color: #000000;\">However, DA = BC and AB = CD (Opposite sides of a rectangle are equal) <\/span><br \/>\n<span style=\"color: #000000;\">Thus AB = BC = CD = DA<\/span><br \/>\n<span style=\"color: #000000;\">ABCD is a rectangle and all the sides are equal. Hence, ABCD is a square.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) diagonal BD bisects \u2220B as well as \u2220D.<br \/>\n<\/strong>Let us join BD<\/span><br \/>\n<span style=\"color: #000000;\">Since ABCD is square,\u00a0AB || CD and\u00a0BC || AD<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394BCD,<\/span><br \/>\n<span style=\"color: #000000;\">BC = CD (Sides of a\u00a0square\u00a0are equal to each other)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220CDB = \u2220CBD (Angles opposite to equal sides are equal)\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000;\">However, \u2220CDB = \u2220ABD (Alternate interior angles as AB || CD)\u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212; (2)<\/span><br \/>\n<span style=\"color: #000000;\">From equations (1) and (2), \u2220CBD = \u2220ABD<\/span><br \/>\n<span style=\"color: #000000;\">Thus, BD bisects \u2220B.<\/span><br \/>\n<span style=\"color: #000000;\">Also, \u2220CBD = \u2220ADB (Alternate interior angles for BC || AD) &#8212;&#8212;&#8212;&#8212; (3)<\/span><br \/>\n<span style=\"color: #000000;\">So, using equations (1) and (3), \u2220ADB =\u00a0\u2220CDB<\/span><br \/>\n<span style=\"color: #000000;\">Hence, BD bisects \u2220D and \u2220B.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4444\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q9.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"359\" height=\"241\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q9.png 383w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q9-300x201.png 300w\" sizes=\"auto, (max-width: 359px) 100vw, 359px\" \/><br \/>\n(i)<\/strong> \u0394 APD \u2245 \u0394 CQB<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii)<\/strong> AP = CQ<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iii)<\/strong> \u0394 AQB \u2245 \u0394 CPD<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iv)<\/strong> AQ = CP<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(v)<\/strong> APCQ is a parallelogram<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Given: ABCD is a\u00a0parallelogram\u00a0and DP = BQ<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) \u0394 APD \u2245 \u0394 CQB<br \/>\n<\/strong>In \u0394APD and \u0394CQB,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADP = \u2220CBQ (Alternate interior angles for BC || AD)<\/span><br \/>\n<span style=\"color: #000000;\">AD = CB (Opposite sides of parallelogram ABCD)<\/span><br \/>\n<span style=\"color: #000000;\">DP = BQ (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394APD \u2245 \u0394CQB (Using\u00a0SAS congruence\u00a0rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) AP = CQ<br \/>\n<\/strong>Since\u00a0\u0394APD \u2245 \u0394CQB,<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AP = CQ (By CPCT)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) \u0394 AQB \u2245 \u0394 CPD<br \/>\n<\/strong>In \u0394AQB and \u0394CPD,<\/span><br \/>\n<span style=\"color: #000000;\">AB = CD (Opposite sides of parallelogram ABCD)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABQ = \u2220CDP (Alternate interior angles\u00a0for AB || CD)<\/span><br \/>\n<span style=\"color: #000000;\">BQ = DP (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394AQB \u2245 \u0394CPD (Using SAS congruence rule)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iv) AQ = CP<br \/>\n<\/strong>Since\u00a0\u0394AQB \u2245 \u0394CPD,<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AQ = CP (CPCT)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(v) APCQ is a parallelogram<br \/>\n<\/strong>From the result obtained in (ii) and (iv), AQ = CP and AP = CQ<\/span><br \/>\n<span style=\"color: #000000;\">Since opposite sides in\u00a0quadrilateral\u00a0APCQ are equal to each other, thus APCQ is a parallelogram.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4446\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q10.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"306\" height=\"207\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q10.png 306w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q10-300x203.png 300w\" sizes=\"auto, (max-width: 306px) 100vw, 306px\" \/><br \/>\n(i) <\/strong>\u0394APB \u2245 \u0394CQD<strong><br \/>\n(ii) <\/strong>AP = CQ<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong><strong>Given:<\/strong> ABCD is a parallelogram and AP \u22a5 DB, CQ \u22a5 DB<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i) \u0394APB \u2245 \u0394CQD<br \/>\n<\/strong>In \u0394APB and \u0394CQD,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220APB = \u2220CQD (Each 90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">AB = CD (Opposite sides of\u00a0parallelogram\u00a0ABCD)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABP = \u2220CDQ (Alternate interior angles\u00a0as AB || CD)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394APB \u2245 \u0394CQD (By AAS congruency)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) AP = CQ<br \/>\n<\/strong>By using the result \u0394APB \u2245 \u0394CQD., we obtain AP = CQ (By CPCT)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11. In \u2206ABC and \u2206DEF, AB = DE, AB || DE, BC \u2013 EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure). <\/strong><strong>Show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4447\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q11.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"304\" height=\"242\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q11.png 304w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q11-300x239.png 300w\" sizes=\"auto, (max-width: 304px) 100vw, 304px\" \/><\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i)<\/strong> quadrilateral ABED is a parallelogram<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii)<\/strong> quadrilateral BEFC is a parallelogram<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iii)<\/strong> AD || CF and AD = CF<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iv)<\/strong> quadrilateral ACFD is a parallelogram<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(v)<\/strong> AC = DF<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(vi<\/strong>) \u2206ABC \u2245 \u2206DEF<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0<\/strong><strong>Given:<\/strong> In \u0394ABC and \u0394DEF, AB = DE, AB || DE, BC = EF and BC || EF.<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i) quadrilateral ABED is a parallelogram<br \/>\n<\/strong>It is given that AB = DE and AB || DE<\/span><br \/>\n<span style=\"color: #000000;\">If one pair of opposite sides of a quadrilateral are equal and parallel to each other, then it will be a\u00a0parallelogram.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, quadrilateral ABED is a parallelogram.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) quadrilateral BEFC is a parallelogram<br \/>\n<\/strong>It is given that BC = EF and BC || EF<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, quadrilateral BEFC\u00a0is a parallelogram.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) AD || CF and AD = CF<br \/>\n<\/strong>As we had observed that ABED and BEFC are parallelograms, therefore<\/span><br \/>\n<span style=\"color: #000000;\">AD = BE\u00a0 and AD || BE (Opposite sides of a parallelogram are equal and parallel)<\/span><br \/>\n<span style=\"color: #000000;\">BE = CF and BE || CF (Opposite sides of a parallelogram are equal and parallel)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, AD = BE = CF and AD ||\u00a0BE || CF<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AD = CF and AD || CF (Lines parallel to the same line are parallel to each other)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) quadrilateral ACFD is a parallelogram<br \/>\n<\/strong>As we had observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and parallel to each other, therefore, it is a parallelogram.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) AC = DF<br \/>\n<\/strong>As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AC || DF and AC = DF<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) \u2206ABC \u2245 \u2206DEF<br \/>\n<\/strong>\u2206ABC and \u2206DEF,<\/span><br \/>\n<span style=\"color: #000000;\">AB = DE (Given)<\/span><br \/>\n<span style=\"color: #000000;\">BC = EF (Given)<\/span><br \/>\n<span style=\"color: #000000;\">AC = DF (Since ACFD is a parallelogram)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2206ABC \u2245 \u2206DEF (By\u00a0SSS congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>12. ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4445\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q12.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"351\" height=\"210\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q12.png 351w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Q12-300x179.png 300w\" sizes=\"auto, (max-width: 351px) 100vw, 351px\" \/><\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i) <\/strong>\u2220A = \u2220B<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(ii) <\/strong>\u2220C = \u2220D<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iii)<\/strong> \u2206ABC \u2245 \u2206BAD<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iv)<\/strong> diagonal AC = diagonal BD<br \/>\n[<strong>Hint:<\/strong> Extend AB and draw a line through C parallel to DA intersecting AB produced at E].<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; <\/strong>Let us join BD and AC in the figure shown below. ADCE is a parallelogram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4448\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans12.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"343\" height=\"190\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans12.png 343w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-8.1-Ans12-300x166.png 300w\" sizes=\"auto, (max-width: 343px) 100vw, 343px\" \/><br \/>\n<strong>(i) \u2220A = \u2220B<br \/>\n<\/strong>AD = CE (Opposite sides of parallelogram AECD are equal)<\/span><br \/>\n<span style=\"color: #000000;\">However, AD = BC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, BC = CE<\/span><br \/>\n<span style=\"color: #000000;\">\u2220CEB = \u2220CBE (Angles opposite to equal sides in a triangle are also equal)<\/span><br \/>\n<span style=\"color: #000000;\">Consider,\u00a0parallel lines\u00a0AD and CE where\u00a0AE is the transversal.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD + \u2220CEB = 180\u00b0 [Co-Interior angles]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD + \u2220CBE = 180\u00b0\u00a0 &#8212;&#8212;&#8212;&#8212;- (1) [Since, \u2220CEB = \u2220CBE]<\/span><br \/>\n<span style=\"color: #000000;\">However, \u2220ABC + \u2220CBE = 180\u00b0 (Linear pair angles) &#8212;&#8212;&#8212;&#8212;- (2)<\/span><br \/>\n<span style=\"color: #000000;\">From Equations (1) and (2), we see that<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD = \u2220ABC<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220A\u00a0= \u2220B<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) \u2220C = \u2220D<br \/>\n<\/strong>AB || CD<\/span><br \/>\n<span style=\"color: #000000;\">\u2220A + \u2220D = 180\u00b0 (Angles on the same side of the\u00a0transversal)<\/span><br \/>\n<span style=\"color: #000000;\">Also, \u2220C + \u2220B = 180\u00b0 (Angles on the same side of the transversal)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220A + \u2220D = \u2220C + \u2220B<\/span><br \/>\n<span style=\"color: #000000;\">However, \u2220A = \u2220B [Using the result obtained in (i)]<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220C = \u2220D<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iii)<\/strong> \u2206ABC \u2245 \u2206BAD<br \/>\nIn \u2206ABC and \u2206BAD,<\/span><br \/>\n<span style=\"color: #000000;\">AB = BA (Common side)<\/span><br \/>\n<span style=\"color: #000000;\">BC = AD (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220B = \u2220A (Proved before)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2206ABC \u2245 \u2206BAD (SAS congruence\u00a0rule)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(iv) diagonal AC = diagonal BD<br \/>\n<\/strong>Since\u00a0\u2206ABC \u2245 \u2206BAD,<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AC = BD (By CPCT)<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 8 (Quadrilaterals)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 8 Quadrilaterals Exercise 8.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 8: Quadrilaterals NCERT Solution Class 9 Maths Ex &#8211; 8.2 Exercise &#8211; 8.1 1. The [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[715,702,186,716,701,5],"class_list":["post-4433","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-8-quadrilaterals-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-8-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Maths\u00a0Chapter - 8 (Quadrilaterals)\u00a0The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter - 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