RS Aggarwal Class 9 Solutions Congruence of Triangles and Inequalities in a Triangle<\/a><\/li>\n<\/ul>\nSome more results based on congruent triangles: \u00a0<\/strong><\/p>\n\nIf two sides of a triangle are unequal, then the longer side has the greater angle opposite to it.<\/li>\n In a triangle, the greater angle has the longer side opposite to it.<\/li>\n Of all the line segments that can be drawn to a given line, from a point not lying on it, the perpendicular line segment is the shortest.<\/li>\n The sum of any two sides of a triangle is greater than its third side.<\/li>\n The difference between any two sides of a triangle is less than its third side.<\/li>\n Exterior angle is greater than one opposite interior angle.<\/li>\n<\/ol>\nCongruent\u00a0Triangles Example Problems With Solutions<\/strong><\/h3>\nExample 1: \u00a0 \u00a0<\/strong>Find the relation between angles in figure. \n \nSolution: \u00a0 \u00a0<\/strong>\u2235 yz > xz > xy \n\u21d2\u2220x > \u2220y > \u2220z. \n(\u2235 Angle opposite to longer side is greater)<\/p>\nExample 2: \u00a0 \u00a0<\/strong>Find the relation between the sides of triangle in figure. \n \nSolution: \u00a0 \u00a0<\/strong>\u2235 \u2220D > \u2220E > \u2220F \n\u2234EF > DF > DE \n{\u2235 side opposite to greater angle is longer}<\/p>\nExample 3: \u00a0 \u00a0<\/strong>Find \u2220ACD then what is the relation between (i) \u2220ACD, \u2220ABC (ii) \u2220ACD & \u2220A \n \nSolution: \u00a0 \u00a0<\/strong>\u2220ACD + 40\u00b0 = 180\u00b0 \u00a0 \u00a0(linear pair) \n\u2220ACD = 140\u00b0 \nalso \u2220A + \u2220B = \u2220ACD \n(exterior angle = sum of opp. interior angles) \n\u21d2 \u2220A + 70\u00b0 = 140\u00b0 \u21d2 \u2220A = 140\u00b0 \u2013 70\u00b0 \n\u21d2 \u2220A = 70\u00b0 \nNow \u2220ACD > \u2220B \n\u2220ACD > \u2220A<\/p>\nExample 4: \u00a0 \u00a0<\/strong>In Fig. \u2220E > \u2220A and \u2220C > \u2220D.\u00a0Prove that AD > EC. \n \nSolution: \u00a0 \u00a0<\/strong>In \u2206ABE, it is given that \n\u2220E > \u2220A \n\u21d2 AB > BE \u00a0 \u00a0 \u00a0 …. (i) \nIn \u2206BCD, it is given that \n\u2220C > \u2220D \n\u21d2 BD > BC \u00a0 \u00a0 \u00a0 …. (ii) \nAdding (i) and (ii), we get \nAB + BD > BE + BC \u21d2AD > EC<\/p>\nExample 5: \u00a0 \u00a0<\/strong>AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that \u2220A > \u2220C. \n \nSolution: \u00a0 \u00a0<\/strong>Draw diagonal AC. \n \nIn \u2206ABC, AB < BC \u00a0 \u00a0{\u2235 AB is smallest} \n\u21d2 \u22203 < \u22201 \u00a0 \u00a0 \u00a0 \u2026\u2026(1) \n{angle opp. to longer side is larger} \nAlso in \u2206ADC \nAD < CD \u2235 CD is longest \n\u21d2 \u22204 < \u22202 \u00a0 \u00a0 \u00a0 \u2026..(2) \nadding equation (1) & (2) \n\u22203 + \u22204 < \u22201 + \u22202 \n\u2220C < \u2220A \nor \u2220A > \u2220C Proved.<\/p>\nExample 6: \u00a0 \u00a0<\/strong>In given figure, PR > PQ and PS bisects \u2220QPR. Prove that \u2220PSR > \u2220PSQ. \n \nSolution: \u00a0 \u00a0<\/strong>In \u2206PQR, PR > PQ \n\u21d2 \u2220Q > \u2220R \u00a0 \u00a0 \u00a0\u2026\u2026(1) \n{angle opposite to longer side is greater} \nand \u22201 = \u22202 \u00a0 \u00a0 (\u2235 PS is \u2220bisector) \u00a0 \u00a0 \u00a0\u2026.(2) \n \nNow for \u2206PQS, \u2220PSR = \u2220Q + \u22201 \u00a0 \u00a0 \u00a0 \u2026.(3) \n{exterior angle = sum of opposite interior angle} \n \n& for \u2206PSR, \u2220PSQ = \u2220R + \u22202 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(4) \nBy equation (1), (2), (3), (4), \u2220PSR > \u2220PSQ \nProved.<\/p>\nExample 7: \u00a0 \u00a0<\/strong>AD, BE and CF, the altitudes of \u2206ABC are equal. Prove that \u2206ABC is an equilateral triangle. \nSolution:<\/strong> \nIn right triangles BCE and BFC, we have \nHyp. BC = Hyp. BC \nBE = CF [Given] \nSo, by RHS criterion of congruence, \n \n\u0394BCE \u2245 \u0394BFC. \n\u21d2 \u2220B = \u2220C \n\u21d2 AC = AB …. (i) \n[\u2235 Sides opposite to equal angles are equal] \nSimilarly, \u0394ABD \u2245 \u0394ABE \n\u21d2 \u2220B = \u2220A \n[\u2235 Corresponding parts of congruent triangles are equal] \n\u21d2 AC = BC …. (ii) \n[\u2235 Sides opposite to equal angles are equal] \nFrom (i) and (ii), we get \nAB = BC = AC \nHence, \u0394ABC is an equilateral triangle.<\/p>\nExample 8: \u00a0 \u00a0<\/strong>In Fig. AD = BC and BD = CA. \n \nProve that \u2220ADB = \u2220BCA \u00a0and \u2220DAB = \u2220CBA. \nSolution:<\/strong> \nIn triangles ABD and ABC, we have \nAD = BC [Given] \nBD = CA [Given] \nand AB = AB [Common] \nSo, by SSS congruence criterion, we have \n\u0394ABD \u2245 \u0394CBA \u21d2 \u2220DAB = \u2220ABC \n[\u2235 corresponding parts of congruent triangles are equal] \n\u21d2 \u2220DAB = \u2220CBA<\/p>\nExample 9: \u00a0 \u00a0<\/strong>In Fig. PQ > PR. QS and RS are the bisectors of \u2220Q and \u2220R respectively.\u00a0Prove that SQ > SR. \n \nSolution:<\/strong> \nIn \u0394PQR, we have \nPQ > PR [Given] \n\u21d2 \u2220PRQ > \u2220PQR \n[\u2235 Angle opp. to larger side of a triangle is greater] \n\u21d2 \\(\\frac { 1 }{ 2 } \\)\u2220PRQ > \\(\\frac { 1 }{ 2 } \\)\u2220PQR \n[\u2235 RS and QS are bisectors of\u00a0\u2220PRQ and \u2220PQR respectively] \n\u21d2 \u2220SRQ > \u2220SQR \n\u21d2 SQ > SR \n[\u2235 Side opp. to greater angle is larger]<\/p>\nExample 10: \u00a0 \u00a0<\/strong>In Fig. \n \nif x > y, show that \u2220M > \u2220N. \nSolution: \u00a0 \u00a0<\/strong> \nWe have, \n\u2220LMN + x\u00ba = 180\u00ba …. (i) \n[Angles of a linear pair] \n\u21d2 \u2220LNM + y\u00ba = 180\u00ba …. (ii) \n[Angles of a linear pair] \n\u2234 \u2220LMN + x\u00ba = \u2220LNM + y\u00ba \nBut x > y. Therefore, \n\u2220LMN < \u2220LNM \n\u21d2 \u2220LNM > \u2220LMN \n\u21d2 LM > LN \n[\u2235 Side opp. to greater angle is larger]<\/p>\nExample 11: \u00a0 <\/strong>\u00a0In Fig. AB > AC. Show that AB > AD. \n \nSolution:<\/strong> \nIn \u0394ABC, we have \nAB > AC [Given] \n\u21d2 \u2220ACB > \u2220ABC …. (i) \n[\u2235 Angle opp. to larger side is greater] \nNow, in \u0394ACD, CD is produced to B, forming an ext \u2220ADB. \n\u21d2 \u2220ADB > \u2220ACD \n[\u2235 Exterior angle of \u0394 is greater than each of interior opp. angle] \n\u21d2 \u2220ADB > \u2220ACB … (ii) \n[\u2234 \u2220ACD = \u2220ACB] \nFrom (i) and (ii), we get \n\u2220ADB > \u2220ABC \n\u21d2 \u2220ADB > \u2220ABD [\u2235 \u2220ABC = \u2220ABD] \n\u21d2 AB > AD \n[\u2235 Side opp. to greater angle is larger]<\/p>\nExample 12: \u00a0 \u00a0<\/strong>Prove that any two sides of a triangle are together greater than twice the median drawn to the third side. \n \nSolution:<\/strong> \nTo prove:<\/strong> AB + AC > 2 AD \nConstruction:<\/strong> Produce AD to E such that AD = DE. Join EC. \nProof:<\/strong> In \u0394’s ADB and EDC, we have \nAD = DE\u00a0 \u00a0 \u00a0 \u00a0[By construction] \nBD = DC\u00a0 \u00a0 \u00a0 \u00a0 [\u2235 D is the mid point of BC] \nand, \u2220ADB = \u2220EDC\u00a0 \u00a0[Vertically opp. angles] \nSo, by SAS criterion of congruence \n\u21d2 \u0394ADB \u2245 \u0394EDC \n\u21d2 AB = EC\u00a0 \u00a0 \u00a0[\u2235 corresponding parts of congruent triangles are equal] \nNow in \u0394AEC, we have \nAC + EC > AE\u00a0 \u00a0 \u00a0 [\u2235 Sum of any two sides of a \u0394 is greater than the third] \n\u21d2 AC + AB > 2 AD \n[\u2235 AD =DE\u00a0\u2234 AE = AD + DE = 2AD and EC = AB]<\/p>\nExample 13: \u00a0 \u00a0<\/strong>In Fig. PQR is a triangle and S is any point in its interior, show that SQ + SR \u00a0< PQ + PR. \n \nSolution:<\/strong> \nGiven:<\/strong> S is any point in the interior of \u0394PQR. \nTo Prove:<\/strong> SQ + SR < PQ + PR \nConstruction:<\/strong> Produce QS to meet PR in T. \nProof:<\/strong> In PQT, we have \nPQ + PT > QT\u00a0 \u00a0 \u00a0[\u2235 Sum of the two sides of a\u00a0\u0394 is greater than the third side] \n\u21d2 PQ + PT > QS + ST\u00a0 \u00a0 \u00a0 …. (i) \n[\u2235 QT = QS + ST] \nIn \u0394RST, we have \nST + TR > SR\u00a0 \u00a0 \u00a0 \u00a0…. (ii) \nAdding (i) and (ii), we get \nPQ + PT + ST + TR > SQ + ST + SR \n\u21d2 PQ + (PT + TR) > SQ + SR \n\u21d2 PQ + PR > SQ + SR \u21d2 SQ + SR < PQ + PR.<\/p>\nExample 14: \u00a0 \u00a0<\/strong>In \u2206PQR S is any point on the side QR.\u00a0Show that PQ + QR + RP > 2 PS. \n \nSolution:<\/strong> \nIn \u0394PQS, we have \nPQ + QS > PS\u00a0 \u00a0 \u00a0 \u00a0 … (i) \n[\u2235 Sum of the two sides of a \u0394 is greater than the third side] \nSimilarly, in \u0394PRS, we have \nRP + RS > PS\u00a0 \u00a0 \u00a0 \u00a0…. (ii) \nAdding (i) and (ii), we get \n(PQ + QS) + (RP + RS) > PS + PS \n\u21d2 PQ + (QS + RS) + RP > 2 PS \n\u21d2 PQ + QR + RP > 2 PS \n[\u2235 QS + RS = QR]<\/p>\nExample 15: \u00a0 \u00a0<\/strong>In Fig. T is a point on side QR of \u2206PQR and S is a point such that RT = ST. \n \nProve that PQ + PR > QS. \nSolution:<\/strong> \nIn \u0394PQR, we have \nPQ + PR > QR \n\u21d2 PQ + PR > QT + RT [\u2235 QR = QT + RT] \n\u21d2 PQ + PR > QT + ST\u00a0 \u00a0 …. (i) \n[\u2235 RT = ST (Given)] \nIn \u0394QST, we have \nQT + ST > QS\u00a0 \u00a0…. (ii) \nFrom (i) and (ii), we get \nPQ + PR > QS.<\/p>\nExample 16: \u00a0 \u00a0<\/strong>Find \u2220OBA in given figure \n \nSolution:<\/strong> \n\u2235 \u2220AOB + 198\u00b0 = 360\u00b0 \n\u2220AOB = 360\u00b0 \u2013 198\u00b0 = 162\u00b0 \nand OA = OB = radius of circle \n\u2220A = \u2220B = x (let) \n\u2234 x + x + 162\u00b0 = 180\u00b0 (a.s.p.) \n2x + 18\u00b0 \nx = 9\u00b0 \n\u2234 \u2220OBA = 9\u00b0.<\/p>\n","protected":false},"excerpt":{"rendered":"How Do You Prove Triangles Are Congruent https:\/\/www.youtube.com\/watch?v=GmS-p1S_t5E Congruent Figures Two figures\/objects are said to be congruent if they are exactly of the same shape and size. The relationship between two congruent figures is called congruence. We use the symbol \u2245 for ‘congruent to’. Congruence among line segments: Two line segments are congruent if they … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[6232,1243,1242,1244,1245,71],"yoast_head":"\nHow Do You Prove Triangles Are Congruent - CBSE Library<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n