Solution:<\/strong><\/p>\n\n\n\n Congruent objects or figures are exact copies of each other or we can say mirror images of each other. The relation of two objects being congruent is called congruence.<\/p>\n\n\n\n Consider Ball 1 and Ball 2. These two balls are congruent.<\/p>\n\n\n\n Ball 1 Ball 2<\/p>\n\n\n\n Now consider the two stars below. Star A and Star B are exactly the same in size, colour and shape. These are congruent stars<\/p>\n\n\n\n Star A Star B<\/p>\n\n\n\n 2. Fill in the blanks:<\/strong><\/p>\n\n\n\n (i) Two line segments are congruent if \u2026\u2026..<\/strong><\/p>\n\n\n\n (ii) Two angles are congruent if \u2026\u2026..<\/strong><\/p>\n\n\n\n (iii) Two square are congruent if \u2026\u2026\u2026<\/strong><\/p>\n\n\n\n (iv) Two rectangles are congruent if \u2026\u2026\u2026<\/strong><\/p>\n\n\n\n (v) Two circles are congruent if \u2026\u2026.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) They are of equal lengths<\/p>\n\n\n\n (ii) Their measures are the same or equal.<\/p>\n\n\n\n (iii) Their sides are equal or they have the same side length<\/p>\n\n\n\n (iv) Their dimensions are same that is lengths are equal and their breadths are also equal.<\/p>\n\n\n\n (v) They have same radii<\/p>\n\n\n\n 3. In Fig. 6, \u2220POQ \u2245\u2220ROS, can we say that \u2220POR \u2245\u2220QOS<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given that<\/p>\n\n\n\n \u2220POQ \u2245\u2220ROS<\/p>\n\n\n\n Also \u2220ROQ \u2245\u2220ROQ<\/p>\n\n\n\n Therefore adding \u2220ROQ to both sides of \u2220POQ \u2245\u2220ROS,<\/p>\n\n\n\n We get, \u2220POQ + \u2220ROQ \u2245\u2220ROQ + \u2220ROS<\/p>\n\n\n\n Therefore, \u2220POR \u2245\u2220QOS<\/p>\n\n\n\n 4. In fig. 7, a = b = c, name the angle which is congruent to \u2220AOC<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n From the figure we have<\/p>\n\n\n\n \u2220 AOB = \u2220 BOC = \u2220 COD<\/p>\n\n\n\n Therefore, \u2220 AOB = \u2220COD<\/p>\n\n\n\n Also, \u2220 AOB + \u2220 BOC = \u2220 BOC + \u2220 COD<\/p>\n\n\n\n \u2220 AOC = \u2220 BOD<\/p>\n\n\n\n Hence, \u2220 BOD \u2245\u2220 AOC<\/p>\n\n\n\n 5. Is it correct to say that any two right angles are congruent? Give reasons to justify your answer.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Two right angles are congruent to each other because they both measure 90o<\/sup>.<\/p>\n\n\n\n We know that two angles are congruent if they have the same measure.<\/p>\n\n\n\n 6. In fig. 8, \u2220AOC \u2245\u2220PYR and \u2220BOC \u2245\u2220QYR. Name the angle which is congruent to \u2220AOB.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given that\u2220AOC \u2245\u2220PYR<\/p>\n\n\n\n Also given that \u2220BOC \u2245\u2220QYR<\/p>\n\n\n\n Now, \u2220AOC = \u2220AOB + \u2220BOC and \u2220PYR =\u2220PYQ +\u2220QYR<\/p>\n\n\n\n By putting the value of \u2220AOC and \u2220PYR in \u2220AOC \u2245\u2220PYR<\/p>\n\n\n\n We get, \u2220AOB + \u2220BOC \u2245\u2220PYQ + \u2220QYR<\/p>\n\n\n\n \u2220AOB \u2245\u2220PYQ (\u2220BOC \u2245\u2220QYR)<\/p>\n\n\n\n Hence, \u2220AOB \u2245\u2220PYQ<\/p>\n\n\n\n 7. Which of the following statements are true and which are false;<\/strong><\/p>\n\n\n\n (i) All squares are congruent.<\/strong><\/p>\n\n\n\n (ii) If two squares have equal areas, they are congruent.<\/strong><\/p>\n\n\n\n (iii) If two rectangles have equal areas, they are congruent.<\/strong><\/p>\n\n\n\n (iv) If two triangles have equal areas, they are congruent.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) False.<\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n All the sides of a square are of equal length. However, different squares can have sides of different lengths. Hence all squares are not congruent.<\/p>\n\n\n\n (ii) True.<\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n Two squares that have the same area will have sides of the same lengths. Hence they will be congruent.<\/p>\n\n\n\n (iii) False<\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n Area of a rectangle = length x breadth<\/p>\n\n\n\n Two rectangles can have the same area. However, the lengths of their sides can vary and hence they are not congruent.<\/p>\n\n\n\n (iv) False<\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n Area of a triangle = 1\/2 x base x height<\/p>\n\n\n\n Two triangles can have the same area but the lengths of their sides can vary and hence they cannot be congruent.<\/p>\n\n\n\n Exercise 16.2 Page No: 16.8<\/p>\n\n\n\n 1. In the following pairs of triangle (Fig. 12 to 15), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) In \u0394 ABC and \u0394 DEF<\/p>\n\n\n\n AB = DE = 4.5 cm (Side)<\/p>\n\n\n\n BC = EF = 6 cm (Side) and<\/p>\n\n\n\n AC = DF = 4 cm (Side)<\/p>\n\n\n\n SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.<\/p>\n\n\n\n Therefore, by SSS criterion of congruence, \u0394ABC \u2245 \u0394DEF<\/p>\n\n\n\n (ii) In \u0394 ACB and \u0394 ADB<\/p>\n\n\n\n AC = AD = 5.5cm (Side)<\/p>\n\n\n\n BC = BD = 5cm (Side) and<\/p>\n\n\n\n AB = AB = 6cm (Side)<\/p>\n\n\n\n SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.<\/p>\n\n\n\n Therefore, by SSS criterion of congruence, \u0394ACB \u2245 \u0394ADB<\/p>\n\n\n\n (iii) In \u0394 ABD and \u0394 FEC,<\/p>\n\n\n\n AB = FE = 5cm (Side)<\/p>\n\n\n\n AD = FC = 10.5cm (Side)<\/p>\n\n\n\n BD = CE = 7cm (Side)<\/p>\n\n\n\n SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.<\/p>\n\n\n\n Therefore, by SSS criterion of congruence, \u0394ABD \u2245 \u0394FEC<\/p>\n\n\n\n (iv) In \u0394 ABO and \u0394 DOC,<\/p>\n\n\n\n AB = DC = 4cm (Side)<\/p>\n\n\n\n AO = OC = 2cm (Side)<\/p>\n\n\n\n BO = OD = 3.5cm (Side)<\/p>\n\n\n\n SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.<\/p>\n\n\n\n Therefore, by SSS criterion of congruence, \u0394ABO \u2245 \u0394ODC<\/p>\n\n\n\n 2. In fig.16, AD = DC and AB = BC<\/strong><\/p>\n\n\n\n (i) Is \u0394ABD \u2245 \u0394CBD?<\/strong><\/p>\n\n\n\n (ii) State the three parts of matching pairs you have used to answer (i).<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) Yes \u0394ABD \u2245\u0394CBD by the SSS criterion.<\/p>\n\n\n\n SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.<\/p>\n\n\n\n Hence \u0394ABD \u2245\u0394CBD<\/p>\n\n\n\n (ii) We have used the three conditions in the SSS criterion as follows:<\/p>\n\n\n\n AD = DC<\/p>\n\n\n\n AB = BC and<\/p>\n\n\n\n DB = BD<\/p>\n\n\n\n 3. In Fig. 17, AB = DC and BC = AD.<\/strong><\/p>\n\n\n\n (i) Is \u0394ABC \u2245 \u0394CDA?<\/strong><\/p>\n\n\n\n (ii) What congruence condition have you used?<\/strong><\/p>\n\n\n\n (iii) You have used some fact, not given in the question, what is that?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) From the figure we have AB = DC<\/p>\n\n\n\n BC = AD<\/p>\n\n\n\n And AC = CA<\/p>\n\n\n\n SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.<\/p>\n\n\n\n Therefore by SSS criterion \u0394ABC \u2245 \u0394CDA<\/p>\n\n\n\n (ii) We have used Side Side Side congruence condition with one side common in both the triangles.<\/p>\n\n\n\n (iii)Yes, we have used the fact that AC = CA.<\/p>\n\n\n\n 4. In \u0394PQR \u2245 \u0394EFD,<\/strong><\/p>\n\n\n\n (i) Which side of \u0394PQR equals ED?<\/strong><\/p>\n\n\n\n (ii) Which angle of \u0394PQR equals angle E?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) PR = ED<\/p>\n\n\n\n Since the corresponding sides of congruent triangles are equal.<\/p>\n\n\n\n (ii) \u2220QPR = \u2220FED<\/p>\n\n\n\n Since the corresponding angles of congruent triangles are equal.<\/p>\n\n\n\n 5. Triangles ABC and PQR are both isosceles with AB = AC and PQ = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?<\/strong><\/p>\n\n\n\n It \u2220B = 50\u00b0, what is the measure of \u2220R?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given that AB = AC in isosceles \u0394ABC<\/p>\n\n\n\n And PQ = PR in isosceles \u0394PQR.<\/p>\n\n\n\n Also given that AB = PQ and QR = BC.<\/p>\n\n\n\n Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)<\/p>\n\n\n\n Hence, \u0394ABC \u2245 \u0394PQR<\/p>\n\n\n\n Now<\/p>\n\n\n\n \u2220ABC = \u2220PQR (Since triangles are congruent)<\/p>\n\n\n\n However, \u0394PQR is isosceles.<\/p>\n\n\n\n Therefore, \u2220PRQ = \u2220PQR = \u2220ABC = 50o<\/sup><\/p>\n\n\n\n 6. ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If \u2220BAC = 40\u00b0 and \u2220BDC = 100\u00b0, then find \u2220ADB.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given ABC and DBC are both isosceles triangles on a common base BC<\/p>\n\n\n\n \u2220BAD = \u2220CAD (corresponding parts of congruent triangles)<\/p>\n\n\n\n \u2220BAD + \u2220CAD = 40o<\/sup>\/ 2<\/p>\n\n\n\n \u2220BAD = 40o<\/sup>\/2 =20o<\/sup><\/p>\n\n\n\n \u2220ABC + \u2220BCA + \u2220BAC = 180o<\/sup> (Angle sum property)<\/p>\n\n\n\n Since \u0394ABC is an isosceles triangle,<\/p>\n\n\n\n \u2220ABC = \u2220BCA<\/p>\n\n\n\n \u2220ABC +\u2220ABC + 40o <\/sup>= 180o<\/sup><\/p>\n\n\n\n 2 \u2220ABC = 180o<\/sup>\u2013 40o<\/sup> = 140o<\/sup><\/p>\n\n\n\n \u2220ABC = 140o<\/sup>\/2 = 70o<\/sup><\/p>\n\n\n\n \u2220DBC + \u2220 BCD + \u2220 BDC = 180o<\/sup> (Angle sum property)<\/p>\n\n\n\n Since \u0394DBC is an isosceles triangle, \u2220 DBC = \u2220BCD<\/p>\n\n\n\n \u2220DBC + \u2220DBC + 100o <\/sup>= 180o<\/sup><\/p>\n\n\n\n 2 \u2220DBC = 180\u00b0\u2013 100o<\/sup> = 80o<\/sup><\/p>\n\n\n\n \u2220DBC = 80o<\/sup>\/2 = 40o<\/sup><\/p>\n\n\n\n In \u0394 BAD,<\/p>\n\n\n\n \u2220ABD + \u2220BAD + \u2220ADB = 180o <\/sup>(Angle sum property)<\/p>\n\n\n\n 30o<\/sup> + 20o<\/sup> + \u2220ADB = 180o<\/sup> (\u2220ADB = \u2220ABC \u2013 \u2220DBC),<\/p>\n\n\n\n \u2220ADB = 180o<\/sup>\u2013 20o<\/sup>\u2013 30o<\/sup><\/p>\n\n\n\n \u2220ADB = 130o<\/sup><\/p>\n\n\n\n 7. \u0394 ABC and \u0394ABD are on a common base AB, and AC = BD and BC = AD as shown in Fig. 18. Which of the following statements is true?<\/strong><\/p>\n\n\n\n (i) \u0394ABC \u2245 \u0394ABD<\/strong><\/p>\n\n\n\n (ii) \u0394ABC \u2245 \u0394ADB<\/strong><\/p>\n\n\n\n (iii) \u0394ABC \u2245 \u0394BAD<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n In \u0394ABC and \u0394BAD we have,<\/p>\n\n\n\n AC = BD (given)<\/p>\n\n\n\n BC = AD (given)<\/p>\n\n\n\n And AB = BA (corresponding parts of congruent triangles)<\/p>\n\n\n\n Therefore by SSS criterion of congruency, \u0394ABC \u2245 \u0394BAD<\/p>\n\n\n\n Therefore option (iii) is true.<\/p>\n\n\n\n 8. In Fig. 19, \u0394ABC is isosceles with AB = AC, D is the mid-point of base BC.<\/strong><\/p>\n\n\n\n (i) Is \u0394ADB \u2245 \u0394ADC?<\/strong><\/p>\n\n\n\n (ii) State the three pairs of matching parts you use to arrive at your answer.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) Given that AB = AC.<\/p>\n\n\n\n Also since D is the midpoint of BC, BD = DC<\/p>\n\n\n\n Also, AD = DA<\/p>\n\n\n\n Therefore by SSS condition,<\/p>\n\n\n\n \u0394ADB \u2245 \u0394ADC<\/p>\n\n\n\n (ii)We have used AB, AC; BD, DC and AD, DA<\/p>\n\n\n\n 9. In fig. 20, \u0394ABC is isosceles with AB = AC. State if \u0394ABC \u2245 \u0394ACB. If yes, state three relations that you use to arrive at your answer.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given that \u0394ABC is isosceles with AB = AC<\/p>\n\n\n\n SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.<\/p>\n\n\n\n \u0394ABC \u2245 \u0394ACBby SSS condition.<\/p>\n\n\n\n Since, ABC is an isosceles triangle, AB = AC and BC = CB<\/p>\n\n\n\n 10. Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does \u2220ABD equal \u2220ACD? Why or why not?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Yes, the two triangles are congruent because given that ABC and DBC have side BC common, AB = BD and AC = CD<\/p>\n\n\n\n Also from the above data we can say<\/p>\n\n\n\n By SSS criterion of congruency, \u0394ABC \u2245 \u0394DBC<\/p>\n\n\n\n No, \u2220ABD and \u2220ACD are not equal because AB not equal to AC<\/p>\n\n\n\n Exercise 16.3 Page No: 16.14<\/p>\n\n\n\n 1. By applying SAS congruence condition, state which of the following pairs (Fig. 28) of triangle are congruent. State the result in symbolic form<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) From the figure we have OA = OC and OB = OD and<\/p>\n\n\n\n \u2220AOB = \u2220COD which are vertically opposite angles.<\/p>\n\n\n\n Therefore by SAS condition, \u0394AOB \u2245\u0394COD<\/p>\n\n\n\n (ii) From the figure we have BD = DC<\/p>\n\n\n\n \u2220ADB = \u2220ADC = 90o<\/sup> and AD = DA<\/p>\n\n\n\n Therefore, by SAS condition, \u0394ADB \u2245\u0394ADC.<\/p>\n\n\n\n (iii) From the figure we have AB = DC<\/p>\n\n\n\n \u2220ABD = \u2220CDB and BD = DB<\/p>\n\n\n\n Therefore, by SAS condition, \u0394ABD \u2245\u0394CBD<\/p>\n\n\n\n (iv) We have BC = QR<\/p>\n\n\n\n ABC = PQR = 90o<\/sup><\/p>\n\n\n\n And AB = PQ<\/p>\n\n\n\n Therefore, by SAS condition, \u0394ABC\u2245 \u0394PQR.<\/p>\n\n\n\n 2. State the condition by which the following pairs of triangles are congruent.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) AB = AD<\/p>\n\n\n\n BC = CD and AC = CA<\/p>\n\n\n\n Therefore by SSS condition, \u0394ABC\u2245 \u0394ADC<\/p>\n\n\n\n (ii) AC = BD<\/p>\n\n\n\n AD = BC and AB = BA<\/p>\n\n\n\n Therefore, by SSS condition, \u0394ABD \u2245 \u0394BAC<\/p>\n\n\n\n (iii) AB = AD<\/p>\n\n\n\n \u2220BAC = \u2220DAC and AC = CA<\/p>\n\n\n\n Therefore by SAS condition, \u0394BAC \u2245 \u0394DAC<\/p>\n\n\n\n (iv) AD = BC<\/p>\n\n\n\n \u2220DAC = \u2220BCA and AC = CA<\/p>\n\n\n\n Therefore, by SAS condition, \u0394ABC \u2245 \u0394ADC<\/p>\n\n\n\n 3.<\/strong> In fig. 30, line segments AB and CD bisect each other at O. Which of the following statements is true?<\/strong><\/p>\n\n\n\n (i) \u0394AOC \u2245 \u0394DOB<\/strong><\/p>\n\n\n\n (ii) \u0394AOC \u2245 \u0394BOD<\/strong><\/p>\n\n\n\n (iii) \u0394AOC \u2245 \u0394ODB<\/strong><\/p>\n\n\n\n State the three pairs of matching parts, you have used to arrive at the answer.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n From the figure we have,<\/p>\n\n\n\n AO = OB<\/p>\n\n\n\n And, CO = OD<\/p>\n\n\n\n Also, AOC = BOD<\/p>\n\n\n\n Therefore, by SAS condition, \u0394AOC \u2245 \u0394BOD<\/p>\n\n\n\n Hence, (ii) statement is true.<\/p>\n\n\n\n 4. Line-segments AB and CD bisect each other at O. AC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles that are given or otherwise known. Are the two triangles congruent? State in symbolic form, which congruence condition do you use?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n We have AO = OB and CO = OD<\/p>\n\n\n\n Since AB and CD bisect each other at 0.<\/p>\n\n\n\n Also \u2220AOC = \u2220BOD<\/p>\n\n\n\n Since they are opposite angles on the same vertex.<\/p>\n\n\n\n Therefore by SAS congruence condition, \u0394AOC \u2245 \u0394BOD<\/p>\n\n\n\n 5. \u0394ABC is isosceles with AB = AC. Line segment AD bisects <\/strong>\u2220A and meets the base BC in D.<\/strong><\/p>\n\n\n\n (i) Is \u0394ADB \u2245 \u0394ADC?<\/strong><\/p>\n\n\n\n (ii) State the three pairs of matching parts used to answer (i).<\/strong><\/p>\n\n\n\n (iii) Is it true to say that BD = DC?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) We have AB = AC (Given)<\/p>\n\n\n\n \u2220BAD = \u2220CAD (AD bisects \u2220BAC)<\/p>\n\n\n\n Therefore by SAS condition of congruence, \u0394ADB \u2245 \u0394ADC<\/p>\n\n\n\n (ii) We have used AB, AC; \u2220BAD = \u2220CAD; AD, DA.<\/p>\n\n\n\n (iii) Now, \u0394ADB\u2245\u0394ADC<\/p>\n\n\n\n Therefore by corresponding parts of congruent triangles<\/p>\n\n\n\n BD = DC.<\/p>\n\n\n\n 6. In Fig. 31, AB = AD and \u2220BAC = \u2220DAC.<\/strong><\/p>\n\n\n\n (i) State in symbolic form the congruence of two triangles ABC and ADC that is true.<\/strong><\/p>\n\n\n\n (ii) Complete each of the following, so as to make it true:<\/strong><\/p>\n\n\n\n (a) \u2220ABC =<\/strong><\/p>\n\n\n\n (b) \u2220ACD =<\/strong><\/p>\n\n\n\n (c) Line segment AC bisects \u2026.. And \u2026\u2026..<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n i) AB = AD (given)<\/p>\n\n\n\n \u2220BAC = \u2220DAC (given)<\/p>\n\n\n\n AC = CA (common)<\/p>\n\n\n\n Therefore by SAS condition of congruency, \u0394ABC \u2245 \u0394ADC<\/p>\n\n\n\n ii) \u2220ABC = \u2220ADC (corresponding parts of congruent triangles)<\/p>\n\n\n\n \u2220ACD = \u2220ACB (corresponding parts of congruent triangles)<\/p>\n\n\n\n Line segment AC bisects \u2220A and \u2220C.<\/p>\n\n\n\n 7. In fig. 32, AB || DC and AB = DC.<\/strong><\/p>\n\n\n\n (i) Is \u0394ACD \u2245 \u0394CAB?<\/strong><\/p>\n\n\n\n (ii) State the three pairs of matching parts used to answer (i).<\/strong><\/p>\n\n\n\n (iii) Which angle is equal to \u2220CAD?<\/strong><\/p>\n\n\n\n (iv) Does it follow from (iii) that AD || BC?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) Yes by SAS condition of congruency, \u0394ACD \u2245 \u0394CAB.<\/p>\n\n\n\n (ii) We have used AB = DC, AC = CA and \u2220DCA = \u2220BAC.<\/p>\n\n\n\n (iii) \u2220CAD = \u2220ACB since the two triangles are congruent.<\/p>\n\n\n\n (iv) Yes this follows from AD parallel to BC as alternate angles are equal. lf alternate angles are equal then the lines are parallel<\/p>\n\n\n\n Exercise 16.4 Page No: 16.19<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/rd-sharma-solutions-for-class-7-maths-chapter-16-congruence-image-1.png\" "\\"RD")
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