![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/rd-sharma-class-9-maths-chapter-10-ex-10-1-questio-1.png\" "\\"RD")
<\/figure>\n\n\n\nSolution<\/strong>:<\/p>\n\n\n\n Sides BA and CA have been produced such that BA = AD and CA = AE.<\/p>\n\n\n\n To prove: DE \u2225 BC<\/p>\n\n\n\n Consider \u25b3 BAC and \u25b3DAE,<\/p>\n\n\n\n BA = AD and CA= AE (Given)<\/p>\n\n\n\n \u2220BAC = \u2220DAE (vertically opposite angles)<\/p>\n\n\n\n By SAS congruence criterion, we have<\/p>\n\n\n\n \u25b3 BAC \u2243 \u25b3 DAE<\/p>\n\n\n\n We know, corresponding parts of congruent triangles are equal<\/p>\n\n\n\n So, BC = DE and \u2220DEA = \u2220BCA, \u2220EDA = \u2220CBA<\/p>\n\n\n\n Now, DE and BC are two lines intersected by a transversal DB s.t.<\/p>\n\n\n\n \u2220DEA=\u2220BCA (alternate angles are equal)<\/p>\n\n\n\n Therefore, DE \u2225 BC. Proved.<\/p>\n\n\n\n Question 2: In a PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Draw a figure based on given instruction,<\/p>\n\n\n\n In \u25b3PQR, PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP respectively (Given)<\/p>\n\n\n\n To prove : LN = MN<\/p>\n\n\n\n As two sides of the triangle are equal, so \u25b3 PQR is an isosceles triangle<\/p>\n\n\n\n PQ = QR and \u2220QPR = \u2220QRP \u2026\u2026. (i)<\/p>\n\n\n\n Also, L and M are midpoints of PQ and QR respectively<\/p>\n\n\n\n PL = LQ = QM = MR = QR\/2<\/p>\n\n\n\n Now, consider \u0394 LPN and \u0394 MRN,<\/p>\n\n\n\n LP = MR<\/p>\n\n\n\n \u2220LPN = \u2220MRN [From (i)]<\/p>\n\n\n\n \u2220QPR = \u2220LPN and \u2220QRP = \u2220MRN<\/p>\n\n\n\n PN = NR [N is midpoint of PR]<\/p>\n\n\n\n By SAS congruence criterion,<\/p>\n\n\n\n \u0394 LPN \u2243 \u0394 MRN<\/p>\n\n\n\n We know, corresponding parts of congruent triangles are equal.<\/p>\n\n\n\n So LN = MN<\/p>\n\n\n\n Proved.<\/p>\n\n\n\n Question 3: In figure, PQRS is a square and SRT is an equilateral triangle. Prove that<\/strong><\/p>\n\n\n\n (i) PT = QT (ii) \u2220 TQR = 150<\/sup><\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given: PQRS is a square and SRT is an equilateral triangle.<\/p>\n\n\n\n To prove:<\/p>\n\n\n\n (i) PT =QT and (ii) \u2220 TQR =15\u00b0<\/p>\n\n\n\n Now,<\/p>\n\n\n\n PQRS is a square:<\/strong><\/p>\n\n\n\n PQ = QR = RS = SP \u2026\u2026 (i)<\/p>\n\n\n\n And \u2220 SPQ = \u2220 PQR = \u2220 QRS = \u2220 RSP = 90o<\/sup><\/p>\n\n\n\n Also, \u25b3 SRT is an equilateral triangle:<\/strong><\/p>\n\n\n\n SR = RT = TS \u2026\u2026.(ii)<\/p>\n\n\n\n And \u2220 TSR = \u2220 SRT = \u2220 RTS = 60\u00b0<\/p>\n\n\n\n From (i) and (ii)<\/p>\n\n\n\n PQ = QR = SP = SR = RT = TS \u2026\u2026(iii)<\/p>\n\n\n\n From figure,<\/p>\n\n\n\n \u2220TSP = \u2220TSR + \u2220 RSP = 60\u00b0 + 90\u00b0 = 150\u00b0 and<\/p>\n\n\n\n \u2220TRQ = \u2220TRS + \u2220 SRQ = 60\u00b0 + 90\u00b0 = 150\u00b0<\/p>\n\n\n\n => \u2220 TSP = \u2220 TRQ = 1500 <\/sup>\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iv)<\/p>\n\n\n\n By SAS congruence criterion, \u0394 TSP \u2243 \u0394 TRQ<\/p>\n\n\n\n We know, corresponding parts of congruent triangles are equal<\/p>\n\n\n\n So, PT = QT<\/p>\n\n\n\n Proved part (i).<\/p>\n\n\n\n Now, consider \u0394 TQR.<\/p>\n\n\n\n QR = TR [From (iii)]<\/p>\n\n\n\n \u0394 TQR is an isosceles triangle.<\/p>\n\n\n\n \u2220 QTR = \u2220 TQR [angles opposite to equal sides]<\/p>\n\n\n\n Sum of angles in a triangle = 180\u2218<\/sup><\/p>\n\n\n\n => \u2220QTR + \u2220 TQR + \u2220TRQ = 180\u00b0<\/p>\n\n\n\n => 2 \u2220 TQR + 150\u00b0 = 180\u00b0 [From (iv)]<\/p>\n\n\n\n => 2 \u2220 TQR = 30\u00b0<\/p>\n\n\n\n => \u2220 TQR = 150<\/sup><\/p>\n\n\n\n Hence proved part (ii).<\/p>\n\n\n\n Question 4: Prove that the medians of an equilateral triangle are equal.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Consider an equilateral \u25b3ABC, and Let D, E, F are midpoints of BC, CA and AB.<\/p>\n\n\n\n Here, AD, BE and CF are medians of \u25b3ABC.<\/p>\n\n\n\n Now,<\/p>\n\n\n\n D is midpoint of BC => BD = DC<\/p>\n\n\n\n Similarly, CE = EA and AF = FB<\/p>\n\n\n\n Since \u0394ABC is an equilateral triangle<\/p>\n\n\n\n AB = BC = CA \u2026..(i)<\/p>\n\n\n\n BD = DC = CE = EA = AF = FB \u2026\u2026\u2026\u2026(ii)<\/p>\n\n\n\n And also, \u2220 ABC = \u2220 BCA = \u2220 CAB = 60\u00b0 \u2026\u2026\u2026.(iii)<\/p>\n\n\n\n Consider \u0394 ABD and \u0394 BCE<\/p>\n\n\n\n AB = BC [From (i)]<\/p>\n\n\n\n BD = CE [From (ii)]<\/p>\n\n\n\n \u2220 ABD = \u2220 BCE [From (iii)]<\/p>\n\n\n\n By SAS congruence criterion,<\/p>\n\n\n\n \u0394 ABD \u2243 \u0394 BCE<\/p>\n\n\n\n => AD = BE \u2026\u2026..(iv)[Corresponding parts of congruent triangles are equal in measure]<\/p>\n\n\n\n Now, consider \u0394 BCE and \u0394 CAF,<\/p>\n\n\n\n BC = CA [From (i)]<\/p>\n\n\n\n \u2220 BCE = \u2220 CAF [From (iii)]<\/p>\n\n\n\n CE = AF [From (ii)]<\/p>\n\n\n\n By SAS congruence criterion,<\/p>\n\n\n\n \u0394 BCE \u2243 \u0394 CAF<\/p>\n\n\n\n => BE = CF \u2026\u2026\u2026\u2026..(v)[Corresponding parts of congruent triangles are equal]<\/p>\n\n\n\n From (iv) and (v), we have<\/p>\n\n\n\n AD = BE = CF<\/p>\n\n\n\n Median AD = Median BE = Median CF<\/p>\n\n\n\n The medians of an equilateral triangle are equal.<\/p>\n\n\n\n Hence proved<\/p>\n\n\n\n Question 5: In a \u0394 ABC, if \u2220A = 120\u00b0 and AB = AC. Find \u2220B and \u2220C.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n To find: \u2220 B and \u2220 C.<\/p>\n\n\n\n Here, \u0394 ABC is an isosceles triangle since AB = AC<\/p>\n\n\n\n \u2220 B = \u2220 C \u2026\u2026\u2026 (i)[Angles opposite to equal sides are equal]<\/p>\n\n\n\n We know, sum of angles in a triangle = 180\u00b0<\/p>\n\n\n\n \u2220 A + \u2220 B + \u2220 C = 180\u00b0<\/p>\n\n\n\n \u2220 A + \u2220 B + \u2220 B= 180\u00b0 (using (i)<\/p>\n\n\n\n 1200<\/sup> + 2\u2220B = 1800<\/sup><\/p>\n\n\n\n 2\u2220B = 1800<\/sup> \u2013 1200<\/sup> = 600<\/sup><\/p>\n\n\n\n \u2220 B = 30o<\/sup><\/p>\n\n\n\n Therefore, \u2220 B = \u2220 C = 30\u2218<\/sup><\/p>\n\n\n\n Question 6: In a \u0394 ABC, if AB = AC and \u2220 B = 70\u00b0, find \u2220 A.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given: In a \u0394 ABC, AB = AC and \u2220B = 70\u00b0<\/p>\n\n\n\n \u2220 B = \u2220 C [Angles opposite to equal sides are equal]<\/p>\n\n\n\n Therefore, \u2220 B = \u2220 C = 70\u2218<\/sup><\/p>\n\n\n\n Sum of angles in a triangle = 180\u2218<\/p>\n\n\n\n \u2220 A + \u2220 B + \u2220 C = 180o<\/sup><\/p>\n\n\n\n \u2220 A + 70o<\/sup> + 70o <\/sup>= 180o<\/sup><\/p>\n\n\n\n \u2220 A = 180o<\/sup> \u2013 140o<\/sup><\/p>\n\n\n\n \u2220 A = 40o<\/sup><\/p>\n\n\n\n Question 1: In figure, it is given that RT = TS, \u2220 1 = 2 \u2220 2 and \u22204 = 2(\u22203). Prove that \u0394RBT \u2245 \u0394SAT.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n In the figure,<\/p>\n\n\n\n RT = TS \u2026\u2026(i)<\/p>\n\n\n\n \u2220 1 = 2 \u2220 2 \u2026\u2026(ii)<\/p>\n\n\n\n And \u2220 4 = 2 \u2220 3 \u2026\u2026(iii)<\/p>\n\n\n\n To prove: \u0394RBT \u2245 \u0394SAT<\/p>\n\n\n\n Let the point of intersection RB and SA be denoted by O<\/p>\n\n\n\n \u2220 AOR = \u2220 BOS [Vertically opposite angles]<\/p>\n\n\n\n or \u2220 1 = \u2220 4<\/p>\n\n\n\n 2 \u2220 2 = 2 \u2220 3 [From (ii) and (iii)]<\/p>\n\n\n\n or \u2220 2 = \u2220 3 \u2026\u2026(iv)<\/p>\n\n\n\n Now in \u0394 TRS, we have RT = TS<\/p>\n\n\n\n => \u0394 TRS is an isosceles triangle<\/p>\n\n\n\n \u2220 TRS = \u2220 TSR \u2026\u2026(v)<\/p>\n\n\n\n But, \u2220 TRS = \u2220 TRB + \u2220 2 \u2026\u2026(vi)<\/p>\n\n\n\n \u2220 TSR = \u2220 TSA + \u2220 3 \u2026\u2026(vii)<\/p>\n\n\n\n Putting (vi) and (vii) in (v) we get<\/p>\n\n\n\n \u2220 TRB + \u2220 2 = \u2220 TSA + \u2220 3<\/p>\n\n\n\n => \u2220 TRB = \u2220 TSA [From (iv)]<\/p>\n\n\n\n Consider \u0394 RBT and \u0394 SAT<\/p>\n\n\n\n RT = ST [From (i)]<\/p>\n\n\n\n \u2220 TRB = \u2220 TSA [From (iv)]<\/p>\n\n\n\n By ASA criterion of congruence, we have<\/p>\n\n\n\n \u0394 RBT \u2245<\/strong> \u0394 SAT<\/p>\n\n\n\n Question 2: Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.<\/strong><\/p>\n\n\n\n Solution:<\/strong> Lines AB and CD Intersect at O<\/p>\n\n\n\n Such that BC \u2225 AD and<\/p>\n\n\n\n BC = AD \u2026\u2026.(i)<\/p>\n\n\n\n To prove : AB and CD bisect at O.<\/p>\n\n\n\n First we have to prove that \u0394 AOD \u2245 \u0394 BOC<\/p>\n\n\n\n \u2220OCB =\u2220ODA [AD||BC and CD is transversal]<\/p>\n\n\n\n AD = BC [from (i)]<\/p>\n\n\n\n \u2220OBC = \u2220OAD [AD||BC and AB is transversal]<\/p>\n\n\n\n By ASA Criterion:<\/p>\n\n\n\n \u0394 AOD \u2245 \u0394 BOC<\/p>\n\n\n\n OA = OB and OD = OC (By c.p.c.t.)<\/p>\n\n\n\n Therefore, AB and CD bisect each other at O.<\/p>\n\n\n\n Hence Proved.<\/p>\n\n\n\n Question 3: BD and CE are bisectors of \u2220 B and \u2220 C of an isosceles \u0394 ABC with AB = AC. Prove that BD = CE.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n \u0394 ABC is isosceles with AB = AC and BD and CE are bisectors of \u2220 B and \u2220 C We have to prove BD = CE. (Given)<\/p>\n\n\n\n Since AB = AC<\/p>\n\n\n\n => \u2220ABC = \u2220ACB \u2026\u2026(i)[Angles opposite to equal sides are equal]<\/p>\n\n\n\n Since BD and CE are bisectors of \u2220 B and \u2220 C<\/p>\n\n\n\n \u2220 ABD = \u2220 DBC = \u2220 BCE = ECA = \u2220B\/2 = \u2220C\/2 \u2026(ii)<\/p>\n\n\n\n Now, Consider \u0394 EBC = \u0394 DCB<\/p>\n\n\n\n \u2220 EBC = \u2220 DCB [From (i)]<\/p>\n\n\n\n BC = BC [Common side]<\/p>\n\n\n\n \u2220 BCE = \u2220 CBD [From (ii)]<\/p>\n\n\n\n By ASA congruence criterion, \u0394 EBC \u2245 \u0394 DCB<\/p>\n\n\n\n Since corresponding parts of congruent triangles are equal.<\/p>\n\n\n\n => CE = BD<\/p>\n\n\n\n or, BD = CE<\/p>\n\n\n\n Hence proved.<\/p>\n\n\n\n Question 1: In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.<\/strong><\/p>\n\n\n\n Solution<\/strong>:<\/p>\n\n\n\n In two right triangles one side and acute angle of one are equal to the corresponding side and angles of the other. (Given)<\/p>\n\n\n\n To prove: Both the triangles are congruent.<\/p>\n\n\n\n Consider two right triangles such that<\/p>\n\n\n\n \u2220 B = \u2220 E = 90o<\/sup> \u2026\u2026.(i)<\/p>\n\n\n\n AB = DE \u2026\u2026.(ii)<\/p>\n\n\n\n \u2220 C = \u2220 F \u2026\u2026(iii)<\/p>\n\n\n\n Here we have two right triangles, \u25b3 ABC and \u25b3 DEF<\/p>\n\n\n\n From (i), (ii) and (iii),<\/p>\n\n\n\n By AAS congruence criterion, we have \u0394 ABC \u2245 \u0394 DEF<\/p>\n\n\n\n Both the triangles are congruent. Hence proved.<\/p>\n\n\n\n Question 2: If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Let ABC be a triangle such that AD is the angular bisector of exterior vertical angle, \u2220EAC and AD \u2225 BC.<\/p>\n\n\n\n\n\n From figure,<\/p>\n\n\n\n \u22201 = \u22202 [AD is a bisector of \u2220 EAC]<\/p>\n\n\n\n \u22201 = \u22203 [Corresponding angles]<\/p>\n\n\n\n and \u22202 = \u22204 [alternative angle]<\/p>\n\n\n\n From above, we have \u22203 = \u22204<\/p>\n\n\n\n This implies, AB = AC<\/p>\n\n\n\n Two sides AB and AC are equal.<\/p>\n\n\n\n => \u0394 ABC is an isosceles triangle.<\/p>\n\n\n\n Question 3: In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Let \u0394 ABC be isosceles where AB = AC and \u2220 B = \u2220 C<\/p>\n\n\n\n Given: Vertex angle A is twice the sum of the base angles B and C. i.e., \u2220 A = 2(\u2220 B + \u2220 C)<\/p>\n\n\n\n \u2220 A = 2(\u2220 B + \u2220 B)<\/p>\n\n\n\n \u2220 A = 2(2 \u2220 B)<\/p>\n\n\n\n \u2220 A = 4(\u2220 B)<\/p>\n\n\n\n Now, We know that sum of angles in a triangle =180\u00b0<\/p>\n\n\n\n \u2220 A + \u2220 B + \u2220 C =180\u00b0<\/p>\n\n\n\n 4 \u2220 B + \u2220 B + \u2220 B = 180\u00b0<\/p>\n\n\n\n 6 \u2220 B =180\u00b0<\/p>\n\n\n\n \u2220 B = 30\u00b0<\/p>\n\n\n\n Since, \u2220 B = \u2220 C<\/p>\n\n\n\n \u2220 B = \u2220 C = 30\u00b0<\/p>\n\n\n\n And \u2220 A = 4 \u2220 B<\/p>\n\n\n\n \u2220 A = 4 x 30\u00b0 = 120\u00b0<\/p>\n\n\n\n Therefore, angles of the given triangle are 30\u00b0 and 30\u00b0 and 120\u00b0.<\/p>\n\n\n\n Question 4: PQR is a triangle in which PQ = PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.<\/strong><\/p>\n\n\n\n Solution:<\/strong> Given that PQR is a triangle such that PQ = PR and S is any point on the side PQ and ST \u2225 QR.<\/p>\n\n\n\n To prove: PS = PT<\/p>\n\n\n\n Since, PQ= PR, so \u25b3PQR is an isosceles triangle.<\/p>\n\n\n\n \u2220 PQR = \u2220 PRQ<\/p>\n\n\n\n Now, \u2220 PST = \u2220 PQR and \u2220 PTS = \u2220 PRQ[Corresponding angles as ST parallel to QR]<\/p>\n\n\n\n Since, \u2220 PQR = \u2220 PRQ<\/p>\n\n\n\n \u2220 PST = \u2220 PTS<\/p>\n\n\n\n In \u0394 PST,<\/p>\n\n\n\n \u2220 PST = \u2220 PTS<\/p>\n\n\n\n \u0394 PST is an isosceles triangle.<\/p>\n\n\n\n Therefore, PS = PT.<\/p>\n\n\n\n Hence proved.<\/p>\n\n\n\n Question 1: In figure, It is given that AB = CD and AD = BC. Prove that \u0394ADC \u2245 \u0394CBA.<\/strong><\/p>\n\n\n\n Solution<\/strong>:<\/p>\n\n\n\n From figure, AB = CD and AD = BC.<\/p>\n\n\n\n To prove: \u0394ADC \u2245 \u0394CBA<\/p>\n\n\n\n Consider \u0394 ADC and \u0394 CBA.<\/p>\n\n\n\n AB = CD [Given]<\/p>\n\n\n\n BC = AD [Given]<\/p>\n\n\n\n And AC = AC [Common side]<\/p>\n\n\n\n So, by SSS congruence criterion, we have<\/p>\n\n\n\n \u0394ADC\u2245\u0394CBA<\/p>\n\n\n\n Hence proved.<\/p>\n\n\n\n Question 2: In a \u0394 PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given: In \u0394 PQR, PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively<\/p>\n\n\n\n To prove: LN = MN<\/p>\n\n\n\n Join L and M, M and N, N and L<\/p>\n\n\n\n We have PL = LQ, QM = MR and RN = NP[Since, L, M and N are mid-points of PQ, QR and RP respectively]<\/p>\n\n\n\n And also PQ = QR<\/p>\n\n\n\n PL = LQ = QM = MR = PN = LR \u2026\u2026.(i)[ Using mid-point theorem]<\/p>\n\n\n\n MN \u2225 PQ and MN = PQ\/2<\/p>\n\n\n\n MN = PL = LQ \u2026\u2026(ii)<\/p>\n\n\n\n Similarly, we have<\/p>\n\n\n\n LN \u2225 QR and LN = (1\/2)QR<\/p>\n\n\n\n LN = QM = MR \u2026\u2026(iii)<\/p>\n\n\n\n From equation (i), (ii) and (iii), we have<\/p>\n\n\n\n PL = LQ = QM = MR = MN = LN<\/p>\n\n\n\n This implies, LN = MN<\/p>\n\n\n\n Hence Proved.<\/p>\n\n\n\n Question 1: ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given: D is the midpoint of BC and PD = DQ in a triangle ABC.<\/p>\n\n\n\n To prove: ABC is isosceles triangle.<\/p>\n\n\n\n In \u25b3BDP and \u25b3CDQ<\/p>\n\n\n\n PD = QD (Given)<\/p>\n\n\n\n BD = DC (D is mid-point)<\/p>\n\n\n\n \u2220BPD = \u2220CQD = 90o<\/sup><\/p>\n\n\n\n By RHS Criterion: \u25b3BDP \u2245 \u25b3CDQ<\/p>\n\n\n\n BP = CQ \u2026 (i) (By CPCT)<\/p>\n\n\n\n In \u25b3APD and \u25b3AQD<\/p>\n\n\n\n PD = QD (given)<\/p>\n\n\n\n AD = AD (common)<\/p>\n\n\n\n APD = AQD = 90 o<\/sup><\/p>\n\n\n\n By RHS Criterion: \u25b3APD \u2245 \u25b3AQD<\/p>\n\n\n\n So, PA = QA \u2026 (ii) (By CPCT)<\/p>\n\n\n\n Adding (i) and (ii)<\/p>\n\n\n\n BP + PA = CQ + QA<\/p>\n\n\n\n AB = AC<\/p>\n\n\n\n Two sides of the triangle are equal, so ABC is an isosceles.<\/p>\n\n\n\n Question 2: ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that \u0394 ABC is isosceles<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n ABC is a triangle in which BE and CF are perpendicular to the sides AC and AB respectively s.t. BE = CF.<\/p>\n\n\n\n To prove: \u0394 ABC is isosceles<\/p>\n\n\n\n In \u0394 BCF and \u0394 CBE,<\/p>\n\n\n\n \u2220 BFC = CEB = 90o<\/sup> [Given]<\/p>\n\n\n\n BC = CB [Common side]<\/p>\n\n\n\n And CF = BE [Given]<\/p>\n\n\n\n By RHS congruence criterion: \u0394BFC \u2245 \u0394CEB<\/p>\n\n\n\n So, \u2220 FBC = \u2220 EBC [By CPCT]<\/p>\n\n\n\n =>\u2220 ABC = \u2220 ACB<\/p>\n\n\n\n AC = AB [Opposite sides to equal angles are equal in a triangle]<\/p>\n\n\n\n Two sides of triangle ABC are equal.<\/p>\n\n\n\n Therefore, \u0394 ABC is isosceles. Hence Proved.<\/p>\n\n\n\n Question 3: If perpendiculars from any point within an angle on its arms are congruent. Prove that it lies on the bisector of that angle.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Consider an angle ABC and BP be one of the arm within the angle.<\/p>\n\n\n\n Draw perpendiculars PN and PM on the arms BC and BA.<\/p>\n\n\n\n In \u0394 BPM and \u0394 BPN,<\/p>\n\n\n\n \u2220 BMP = \u2220 BNP = 90\u00b0 [given]<\/p>\n\n\n\n BP = BP [Common side]<\/p>\n\n\n\n MP = NP [given]<\/p>\n\n\n\n By RHS congruence criterion: \u0394BPM\u2245\u0394BPN<\/p>\n\n\n\n So, \u2220 MBP = \u2220 NBP [ By CPCT]<\/p>\n\n\n\n BP is the angular bisector of \u2220ABC.<\/p>\n\n\n\n Hence proved<\/p>\n\n\n\n Question 1: In \u0394 ABC, if \u2220 A = 40\u00b0 and \u2220 B = 60\u00b0. Determine the longest and shortest sides of the triangle.<\/strong><\/p>\n\n\n\n Solution:<\/strong> In \u0394 ABC, \u2220 A = 40\u00b0 and \u2220 B = 60\u00b0<\/p>\n\n\n\n We know, sum of angles in a triangle = 180\u00b0<\/p>\n\n\n\n \u2220 A + \u2220 B + \u2220 C = 180\u00b0<\/p>\n\n\n\n 40\u00b0 + 60\u00b0 + \u2220 C = 180\u00b0<\/p>\n\n\n\n \u2220 C = 180\u00b0 \u2013 100\u00b0 = 80\u00b0<\/p>\n\n\n\n \u2220 C = 80\u00b0<\/p>\n\n\n\n Now, 40\u00b0 < 60\u00b0 < 80\u00b0<\/p>\n\n\n\n => \u2220 A < \u2220 B < \u2220 C<\/p>\n\n\n\n => \u2220 C is greater angle and \u2220 A is smaller angle.<\/p>\n\n\n\n Now, \u2220 A < \u2220 B < \u2220 C<\/p>\n\n\n\n We know, side opposite to greater angle is larger and side opposite to smaller angle is smaller.<\/p>\n\n\n\n Therefore, BC < AC < AB<\/p>\n\n\n\n AB is longest and BC is shortest side.<\/p>\n\n\n\n Question 2: In a \u0394 ABC, if \u2220 B = \u2220 C = 45\u00b0, which is the longest side?<\/strong><\/p>\n\n\n\n Solution:<\/strong> In \u0394 ABC, \u2220 B = \u2220 C = 45\u00b0<\/p>\n\n\n\n Sum of angles in a triangle = 180\u00b0<\/p>\n\n\n\n \u2220 A + \u2220 B + \u2220 C = 180\u00b0<\/p>\n\n\n\n \u2220 A + 45\u00b0 + 45\u00b0 = 180\u00b0<\/p>\n\n\n\n \u2220 A = 180\u00b0 \u2013 (45\u00b0 + 45\u00b0) = 180\u00b0 \u2013 90\u00b0 = 90\u00b0<\/p>\n\n\n\n \u2220 A = 90\u00b0<\/p>\n\n\n\n => \u2220 B = \u2220 C < \u2220 A<\/p>\n\n\n\n Therefore, BC is the longest side.<\/p>\n\n\n\n Question 3: In \u0394 ABC, side AB is produced to D so that BD = BC. If \u2220 B = 60\u00b0 and \u2220 A = 70\u00b0.<\/strong><\/p>\n\n\n\n Prove that: (i) AD > CD (ii) AD > AC<\/strong><\/p>\n\n\n\n Solution:<\/strong> In \u0394 ABC, side AB is produced to D so that BD = BC.<\/p>\n\n\n\n \u2220 B = 60\u00b0, and \u2220 A = 70\u00b0<\/p>\n\n\n\n To prove: (i) AD > CD (ii) AD > AC<\/p>\n\n\n\n Construction: Join C and D<\/p>\n\n\n\n We know, sum of angles in a triangle = 180\u00b0<\/p>\n\n\n\n \u2220 A + \u2220 B + \u2220 C = 180\u00b0<\/p>\n\n\n\n 70\u00b0 + 60\u00b0 + \u2220 C = 180\u00b0<\/p>\n\n\n\n \u2220 C = 180\u00b0 \u2013 (130\u00b0) = 50\u00b0<\/p>\n\n\n\n \u2220 C = 50\u00b0<\/p>\n\n\n\n \u2220 ACB = 50\u00b0 \u2026\u2026(i)<\/p>\n\n\n\n And also in \u0394 BDC<\/p>\n\n\n\n \u2220 DBC =180\u00b0 \u2013 \u2220 ABC = 180 \u2013 60\u00b0 = 120\u00b0[\u2220DBA is a straight line]<\/p>\n\n\n\n and BD = BC [given]<\/p>\n\n\n\n \u2220 BCD = \u2220 BDC [Angles opposite to equal sides are equal]<\/p>\n\n\n\n Sum of angles in a triangle =180\u00b0<\/p>\n\n\n\n \u2220 DBC + \u2220 BCD + \u2220 BDC = 180\u00b0<\/p>\n\n\n\n 120\u00b0 + \u2220 BCD + \u2220 BCD = 180\u00b0<\/p>\n\n\n\n 120\u00b0 + 2\u2220 BCD = 180\u00b0<\/p>\n\n\n\n 2\u2220 BCD = 180\u00b0 \u2013 120\u00b0 = 60\u00b0<\/p>\n\n\n\n \u2220 BCD = 30\u00b0<\/p>\n\n\n\n \u2220 BCD = \u2220 BDC = 30\u00b0 \u2026.(ii)<\/p>\n\n\n\n Now, consider \u0394 ADC.<\/p>\n\n\n\n \u2220 DAC = 70\u00b0 [given]<\/p>\n\n\n\n \u2220 ADC = 30\u00b0 [From (ii)]<\/p>\n\n\n\n \u2220 ACD = \u2220 ACB+ \u2220 BCD = 50\u00b0 + 30\u00b0 = 80\u00b0 [From (i) and (ii)]<\/p>\n\n\n\n Now, \u2220 ADC < \u2220 DAC < \u2220 ACD<\/p>\n\n\n\n AC < DC < AD[Side opposite to greater angle is longer and smaller angle is smaller]<\/p>\n\n\n\n AD > CD and AD > AC<\/p>\n\n\n\n Hence proved.<\/p>\n\n\n\n Question 4: Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Lengths of sides are 2 cm, 3 cm and 7 cm.<\/p>\n\n\n\n A triangle can be drawn only when the sum of any two sides is greater than the third side.<\/p>\n\n\n\n So, let\u2019s check the rule.<\/p>\n\n\n\n 2 + 3 \u226f 7 or 2 + 3 < 7<\/p>\n\n\n\n 2 + 7 > 3<\/p>\n\n\n\n and 3 + 7 > 2<\/p>\n\n\n\n Here 2 + 3 \u226f 7<\/p>\n\n\n\n So, the triangle does not exit.<\/p>\n\n\n\n Question 1: In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n In two congruent triangles ABC and DEF, if AB = DE and BC = EF, then<\/p>\n\n\n\n \u2220A = \u2220D, \u2220B = \u2220 E and \u2220 C = \u2220F<\/p>\n\n\n\n Question 2: In two triangles ABC and DEF, it is given that \u2220A = \u2220D, \u2220B = \u2220 E and \u2220 C = \u2220F. Are the two triangles necessarily congruent?<\/strong><\/p>\n\n\n\n Reason: Two triangles are not necessarily congruent, because we know only angle-angle-angle (AAA) criterion. This criterion can produce similar but not congruent triangles.<\/p>\n\n\n\n Question 3: If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, C = 75o<\/sup>, DE = 2.5 cm, DF = 5 cm and D = 75o<\/sup>. Are two triangles congruent?<\/strong><\/p>\n\n\n\n Solution<\/strong>: Yes.<\/p>\n\n\n\n Reason: Given triangles are congruent as AC = DE = 2.5 cm, BC = DF = 5 cm and<\/p>\n\n\n\n \u2220D = \u2220C = 75o<\/sup>.<\/p>\n\n\n\n By SAS theorem triangle ABC is congruent to triangle EDF.<\/p>\n\n\n\n Question 4: In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?<\/strong><\/p>\n\n\n\n Solution:<\/strong> Yes.<\/p>\n\n\n\n Reason: Given triangles are congruent as<\/p>\n\n\n\n AB = AD<\/p>\n\n\n\n BC = CD and<\/p>\n\n\n\n AC [ common side]<\/p>\n\n\n\n By SSS theorem triangle ABC is congruent to triangle ADC.<\/p>\n\n\n\n Question 5: In triangles ABC and CDE, if AC = CE, BC = CD, \u2220A = 60o<\/sup>, \u2220C = 30 o<\/sup> and \u2220D = 90 o<\/sup>. Are two triangles congruent?<\/strong><\/p>\n\n\n\n Solution: <\/strong>Yes.<\/p>\n\n\n\n Reason: Given triangles are congruent<\/p>\n\n\n\n Here AC = CE<\/p>\n\n\n\n BC = CD<\/p>\n\n\n\n \u2220B = \u2220D = 90\u00b0<\/p>\n\n\n\n By SSA criteria triangle ABC is congruent to triangle CDE.<\/p>\n\n\n\n Question 6: ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.<\/strong><\/p>\n\n\n\n Solution:<\/strong> ABC is an isosceles triangle (given)<\/p>\n\n\n\n AB = AC (given)<\/p>\n\n\n\n BE and CF are two medians (given)<\/p>\n\n\n\n To prove: BE = CF<\/p>\n\n\n\n In \u25b3CFB and \u25b3BEC<\/p>\n\n\n\n CE = BF (Since, AC = AB = AC\/2 = AB\/2 = CE = BF)<\/p>\n\n\n\n BC = BC (Common)<\/p>\n\n\n\n \u2220ECB = \u2220FBC (Angle opposite to equal sides are equal)<\/p>\n\n\n\n By SAS theorem: \u25b3CFB \u2245 \u25b3BEC<\/p>\n\n\n\n So, BE = CF (By c.p.c.t)<\/p>\n\n\n\n RD Sharma Solutions for Class 9 Maths Chapter 10\u2013Congruent Triangles<\/p>\n\n\n\n Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 10\u2013Congruent Triangles PDF<\/a><\/p>\n\n\n\n RD Sharma i![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/rd-sharma-class-9-maths-chapter-10-ex-10-1-questio-2.png\" "\\"RD")
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\n\n\n\nExercise 10.2<\/h4>\n\n\n\n
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\n\n\n\nExercise 10.3<\/h4>\n\n\n\n
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\n\n\n\nExercise 10.4<\/h4>\n\n\n\n
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\n\n\n\nExercise 10.5<\/h4>\n\n\n\n
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\n\n\n\nExercise 10.6 Page No: 10.66<\/h4>\n\n\n\n
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\n\n\n\nExercise VSAQs<\/h4>\n\n\n\n
Solution:<\/strong> No.<\/p>\n\n\n\nRD Sharma Solutions for Class 9 Maths Chapter 10: Download PDF<\/strong><\/h2>\n\n\n\n
Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n
About RD Sharma<\/h2>\n\n\n\n