{"id":684,"date":"2018-08-18T05:13:14","date_gmt":"2018-08-18T05:13:14","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=684"},"modified":"2018-08-18T09:07:27","modified_gmt":"2018-08-18T09:07:27","slug":"similarity-of-triangles","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/similarity-of-triangles\/","title":{"rendered":"Criteria For Similarity Of Triangles"},"content":{"rendered":"<h2><strong>Criteria For Similarity Of <a href=\"https:\/\/cbselibrary.com\/classification-of-triangles\/\">Triangles<\/a><\/strong><\/h2>\n<p><strong>AAA similarity criterion:<\/strong> If in two triangles, corresponding angles are equal, then the triangles are similar.<br \/>\n<strong>AA Similarity criterion:<\/strong> If in two triangles, two angles of one triangle are respectively equal the two angles of the other triangle, then the two triangles are similar.<br \/>\n<strong>SSS Similarity criterion:<\/strong> If in two triangles, corresponding sides are in the same ratio, then the two triangles are similar.<br \/>\n<strong>SAS similarity criterion:<\/strong>\u00a0If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the triangles are similar.<br \/>\n<strong>Equiangular Triangles:<\/strong><br \/>\nTwo triangles are said to be equiangular, if their corresponding angles are equal.<br \/>\nIf two triangles are equiangular, then they are similar.<br \/>\nTwo triangles ABC and DEF such that<br \/>\n\u2220A = \u2220D, \u2220B = \u2220E and \u2220C = \u2220F.<br \/>\nThen \u2206ABC ~ \u2206DEF and<br \/>\n\\(\\frac{AB}{DE}=\\frac{BC}{EF}=\\frac{AC}{DF}\\)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8455\/29026999176_959886ccc7_o.png\" alt=\"similarity-of-triangles-1\" width=\"267\" height=\"130\" \/><\/p>\n<p><b>Read More:<\/b><\/p>\n<ul>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/angle-sum-property-of-triangle\/\"><span style=\"font-weight: 400;\">Angle Sum Property of a Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/median-altitude-of-triangle\/\"><span style=\"font-weight: 400;\">Median and Altitude of a Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/find-angle-of-isosceles-triangle\/\"><span style=\"font-weight: 400;\">The Angle of An Isosceles Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/areas-of-two-similar-triangles\/\"><span style=\"font-weight: 400;\">Areas of Two Similar Triangles<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/area-of-a-triangle\/\"><span style=\"font-weight: 400;\">Area of A Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/congruent-triangles\/\"><span style=\"font-weight: 400;\">To Prove Triangles Are Congruent<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/construction-of-triangles-with-compass-and-protractor\/\"><span style=\"font-weight: 400;\">Construction of an Equilateral Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/cbselibrary.com\/classification-of-triangles\/\"><span style=\"font-weight: 400;\">Classification of Triangles<\/span><\/a><\/li>\n<\/ul>\n<h2><strong>Criteria For Similarity Of Triangles With Examples<\/strong><\/h2>\n<p><strong>Example 1: \u00a0 \u00a0<\/strong>In figure, find \u2220L.<br \/>\n<img loading=\"lazy\" class=\"\" src=\"https:\/\/c2.staticflickr.com\/8\/7578\/28954563662_48dc14e401_o.png\" alt=\"similarity-of-triangles-2\" width=\"287\" height=\"195\" \/><br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>In \u2206ABC and \u2206LMN,<br \/>\n\\( \\frac{AB}{LM}=\\frac{4.4}{11}=\\frac{2}{5}\\)<br \/>\n\\( \\frac{BC}{MN}=\\frac{4}{10}=\\frac{2}{5} \\)<br \/>\n\\( \\frac{CA}{NL}=\\frac{3.6}{9}=\\frac{2}{5} \\)<br \/>\n\\( \\Rightarrow \\frac{AB}{LM}=\\frac{BC}{MN}=\\frac{CA}{NL} \\)<br \/>\n\u2206ABC ~ \u2206LMN (SSS similarity)<br \/>\n\u2220L = \u2220A<br \/>\n= 180\u00ba \u2013 \u2220B \u2013 \u2220C<br \/>\n= 180\u00ba \u2013 50\u00ba \u2013 70\u00ba = 60\u00ba<br \/>\n\u2220L = 60\u00ba<\/p>\n<p><strong>Example 2: \u00a0 \u00a0<\/strong>Examine each pair of triangles in Figure, and state which pair of triangles are similar. Also, state the\u00a0similarity criterion used by you for answering the question and write the similarity relation in symbolic form.<br \/>\n<img loading=\"lazy\" class=\"\" src=\"https:\/\/c1.staticflickr.com\/9\/8057\/28983817331_2f6a94caf2_o.png\" alt=\"similarity-of-triangles-3\" width=\"251\" height=\"143\" \/><br \/>\nFigure (i)<br \/>\n<img loading=\"lazy\" class=\"\" src=\"https:\/\/c1.staticflickr.com\/9\/8015\/29027208426_452aaaa441_o.png\" alt=\"similarity-of-triangles-4\" width=\"277\" height=\"184\" \/><br \/>\nFigure (ii)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8029\/28983817151_a6313610fc_o.png\" alt=\"similarity-of-triangles-5\" width=\"283\" height=\"172\" \/><br \/>\nFigure (iii)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8360\/29027208186_0b0d06edd4_o.png\" alt=\"similarity-of-triangles-6\" width=\"277\" height=\"191\" \/><br \/>\nFigure (iv)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8101\/28983816881_3735ab826f_o.png\" alt=\"similarity-of-triangles-7\" width=\"273\" height=\"146\" \/><br \/>\nFigure (v)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8422\/29027207946_16124915ae_o.png\" alt=\"similarity-of-triangles-8\" width=\"301\" height=\"119\" \/><br \/>\nFigure (vi)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8456\/28983816671_f89b3615ae_o.png\" alt=\"similarity-of-triangles-9\" width=\"335\" height=\"154\" \/><br \/>\nFigure (vii)<br \/>\n<strong>Sol. \u00a0 \u00a0\u00a0(i)<\/strong>\u00a0\u2220A = \u2220Q, \u2220B = \u2220P and \u2220C = \u2220R.<br \/>\n\u2234\u00a0\u2206ABC ~ \u2206QPR \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(AAA-similarity)<br \/>\n<strong>(ii)<\/strong> In triangle PQR and DEF, we observe that<br \/>\n\\(\\frac{PQ}{DE}=\\frac{QR}{EF}=\\frac{PR}{DF}=\\frac{1}{2} \\)<br \/>\nTherefore, by SSS-criterion of similarity, we have<br \/>\n\u2206PQR ~ \u2206DEF<br \/>\n<strong>(iii)<\/strong> SAS-similarity is not satisfied as included angles are not equal.<br \/>\n<strong>(iv)<\/strong>\u00a0\u2206CAB ~ \u2206QRP \u00a0 \u00a0 \u00a0 \u00a0 (SAS-similarity), as<br \/>\n\\(\\frac{CA}{QR}=\\frac{CB}{QP} \\) \u00a0 and \u2220C = \u2220Q.<br \/>\n<strong>(v)<\/strong> In \u2206\u2019s ABC and DEF, we have<br \/>\n\u2220A = \u2220D = 80\u00ba<br \/>\n\\(\\text{But, }\\frac{AB}{DE}\\ne \\frac{AC}{DF} \\) \u00a0 \u00a0 \u00a0 \u00a0[\u2235\u00a0 AC is not given]<br \/>\nSo, by SAS-criterion of similarity these two triangles are not similar.<br \/>\n<strong>(vi)<\/strong> In \u2206\u2019s DEF and MNP, we have<br \/>\n\u2220D = \u2220M = 70\u00ba<br \/>\n\u2220E = \u2220N = 80\u00ba \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [\u2235\u00a0 \u2220N = 180\u00ba \u2013 \u2220M \u2013 \u2220P = 180\u00ba \u2013 70\u00ba \u2013 30\u00ba = 80\u00ba]<br \/>\nSo, by AA-criterion of similarity<br \/>\n\u2206DEF ~ \u2206MNP.<br \/>\n<strong>(vii)<\/strong> FE = 2 cm, FD = 3 cm, ED = 2.5 cm<br \/>\nPQ = 4 cm, PR = 6 cm, QR = 5 cm<br \/>\n\u2234\u00a0\u2206FED ~ \u2206PQR \u00a0 \u00a0 \u00a0 \u00a0(SSS-similarity)<\/p>\n<p><strong>Example 3: \u00a0 \u00a0<\/strong>In figure, QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Find AQ.<br \/>\n<strong>Sol. \u00a0 \u00a0\u00a0<\/strong>In triangles AOQ and BOP, we have<br \/>\n\u2220OAQ = \u2220OBP \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Each equal to 90\u00ba]<br \/>\n\u2220AOQ = \u2220BOP \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Vertically opposite angles]<br \/>\nTherefore, by AA-criterion of similarity<br \/>\n\u2206AOQ ~ \u2206BOP<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8409\/28955032142_7f6f2a3063_o.png\" alt=\"similarity-of-triangles-10\" width=\"181\" height=\"164\" \/><br \/>\n\\( \\Rightarrow \\frac{AO}{BO}=\\frac{OQ}{OP}=\\frac{AO}{BP} \\)<br \/>\n\\( \\Rightarrow \\frac{AO}{BO}=\\frac{AQ}{BP}\\Rightarrow \\frac{10}{6}=\\frac{AQ}{9} \\)<br \/>\n\\( \\Rightarrow AQ=\\frac{10\\times 9}{6}=15\\text{ }cm \\)<\/p>\n<p><strong>Example 4: \u00a0 \u00a0\u00a0<\/strong>In figure, \u2206ACB ~ \u2206APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, find CA and AQ.<br \/>\n<strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 We have, \u2206ACB ~ \u2206APQ<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8210\/28774133630_d7e5d9869b_o.png\" alt=\"similarity-of-triangles-11\" width=\"170\" height=\"139\" \/><br \/>\n\\( \\Rightarrow \\frac{AC}{AP}=\\frac{CB}{PQ}=\\frac{AB}{AQ} \u00a0\\)<br \/>\n\\( \\Rightarrow \\frac{AC}{AP}=\\frac{CB}{PQ}\\text{\u00a0 and\u00a0\u00a0 }\\frac{CB}{PQ}=\\frac{AB}{AQ} \\)<br \/>\n\\( \\Rightarrow \\frac{AC}{2.8}=\\frac{8}{4}\\text{\u00a0 and\u00a0\u00a0 }\\frac{8}{4}=\\frac{6.5}{AQ} \u00a0\\)<br \/>\n\\( \\Rightarrow \\frac{AC}{2.8}=2\\text{\u00a0\u00a0 and\u00a0\u00a0 }\\frac{6.5}{AQ}=2 \\)<br \/>\nAC = (2 \u00d7 2.8) cm = 5.6 cm and \u00a0 \u00a0\\( AQ=\\frac{6.5}{2}\\text{ }cm\\text{ }=\\text{ }3.25\\text{ }cm \\)<\/p>\n<p><strong>Example 5: \u00a0 \u00a0\u00a0<\/strong>The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB.<br \/>\n<strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0Since the ratio of the corresponding sides of similar triangles is same as the ratio of their perimeters.<br \/>\n\u2234\u00a0\u2206ABC ~ \u2206PQR<br \/>\n\\( \\Rightarrow \\frac{AB}{PQ}=\\frac{BC}{QR}=\\frac{AC}{PR}=\\frac{36}{24} \\)<br \/>\n\\( \\Rightarrow \\frac{AB}{PQ}=\\frac{36}{24}\\Rightarrow \\frac{AB}{10}=\\frac{36}{24} \\)<br \/>\n\\( \\Rightarrow AB=\\frac{36\\times 10}{24}cm\\text{ }=\\text{ }15\\text{ }cm \\)<\/p>\n<p><strong>Example 6: \u00a0 \u00a0\u00a0<\/strong>In figure, \u2220CAB = 90\u00ba and AD \u22a5\u00a0BC. If AC = 75 cm, AB = 1 m and BD = 1.25 m, find AD.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0\u00a0We have,<br \/>\nAB = 1 m = 100 cm, AC = 75 cm and BD = 125 cm<br \/>\nIn \u2206BAC and \u2206BDA, we have<br \/>\n\u2220BAC = \u2220BDA \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Each equal to 90\u00ba]<br \/>\nand, \u2220B = \u2220B<br \/>\nSo, by AA-criterion of similarity, we have<br \/>\n\u2206BAC ~ \u2206BDA<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8294\/28439524284_8356f0556b_o.png\" alt=\"similarity-of-triangles-12\" width=\"162\" height=\"167\" \/><br \/>\n\\( \\Rightarrow \\frac{BA}{BD}=\\frac{AC}{AD} \\)<br \/>\n\\( \\Rightarrow \\frac{100}{125}=\\frac{75}{AD} \\)<br \/>\n\\( \\Rightarrow AD=\\frac{125\\times 75}{100}cm\\text{ }=\\text{ }93.75\\text{ }cm \\)<\/p>\n<p><strong>Example 7: \u00a0 \u00a0\u00a0<\/strong>In figure, if \u2220A = \u2220C, then prove that \u2206AOB ~ \u2206COD.<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8017\/28439524194_2acfd99cdd_o.png\" alt=\"similarity-of-triangles-13\" width=\"145\" height=\"205\" \/><br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0\u00a0In triangles AOB and COD, we have<br \/>\n\u2220A = \u2220C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Given]<br \/>\nand, \u22201 = \u22202 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Vertically opposite angles]<br \/>\nTherefore, by AA-criterion of similarity, we have<br \/>\n\u2206AOB ~ \u2206COD<\/p>\n<p><strong>Example 8: \u00a0 \u00a0\u00a0<\/strong>In figure, \u00a0\\( \\frac{AO}{OC}=\\frac{BO}{OD}=\\frac{1}{2} \\) \u00a0 and AB = 5 cm. Find the value of DC.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 In \u2206AOB and \u2206COD, we have<br \/>\n\u2206AOB = \u2206COD \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Vertically opposite angles]<br \/>\n\\( \\frac{AO}{OC}=\\frac{BO}{OD} \\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Given]<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8331\/28774133460_f4522f3dc7_o.png\" alt=\"similarity-of-triangles-14\" width=\"239\" height=\"180\" \/><br \/>\nSo, by SAS-criterion of similarity, we have<br \/>\n\u2206AOB ~ \u2206COD<br \/>\n\\( \\Rightarrow \\frac{AO}{OC}=\\frac{BO}{OD}=\\frac{AB}{DC} \\)<br \/>\n\\( \\Rightarrow \\frac{1}{2}=\\frac{5}{DC} \\) \u00a0 \u00a0 \u00a0 \u00a0[\u2235\u00a0 \u00a0AB = 5 cm]<br \/>\n\u21d2 DC = 10 cm<\/p>\n<p><strong>Example 9: \u00a0 \u00a0\u00a0<\/strong>In figure, considering triangles BEP and CPD, prove that BP \u00d7 PD = EP \u00d7 PC.<br \/>\n<strong>Sol. \u00a0 \u00a0Given:<\/strong> A \u2206ABC in which BD \u22a5\u00a0AC and CE \u22a5\u00a0AB and BD and CE intersect at P.<br \/>\n<strong>To Prove:<\/strong> BP \u00d7 PD = EP \u00d7 PC<br \/>\n<strong>Proof:<\/strong> In \u2206EPB and \u2206DPC, we have<br \/>\n\u2220PEB = \u2220PDC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Each equal to 90\u00ba]<br \/>\n\u2220EPB = \u2220DPC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Vertically opposite angles]<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8284\/28955031792_123347c71b_o.png\" alt=\"similarity-of-triangles-15\" width=\"179\" height=\"144\" \/><br \/>\nThus, by AA-criterion of similarity, we have<br \/>\n\u2206EPB ~ \u2206DPC<br \/>\n\\(\\frac{EP}{DP}=\\frac{PB}{PC}\\)<br \/>\n\u21d2 BP \u00d7 PD = EP \u00d7 PC<\/p>\n<p><strong>Example 10: \u00a0 \u00a0\u00a0<\/strong>D is a point on the side BC of \u2206ABC such that \u2220ADC = \u2220BAC.<br \/>\nProve that \u00a0\\(\\frac{CA}{CD}=\\frac{CB}{CA}\\) \u00a0 or, CA<sup>2<\/sup> = CB \u00d7 CD.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> In \u2206ABC and \u2206DAC, we have<br \/>\n\u2220ADC = \u2220BAC and \u2220C = \u2220C<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8314\/28955031732_bed868c46f_o.png\" alt=\"similarity-of-triangles-16\" width=\"189\" height=\"165\" \/><br \/>\nTherefore, by AA-criterion of similarity, we have<br \/>\n\u2206ABC ~ \u2206DAC<br \/>\n\\( \\Rightarrow \\frac{AB}{DA}=\\frac{BC}{AC}=\\frac{AC}{DC} \\)<br \/>\n\\( \\Rightarrow \\frac{CB}{CA}=\\frac{CA}{CD} \\)<\/p>\n<p><strong>Example 11: \u00a0 \u00a0\u00a0<\/strong>P and Q are points on sides AB and AC respectively of \u2206ABC. If AP = 3 cm, PB = 6cm. AQ = 5 cm and QC = 10 cm, show that BC = 3PQ.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> We have,<br \/>\nAB = AP + PB = (3 + 6) cm = 9 cm<br \/>\nand, AC = AQ + QC = (5 + 10) cm = 15 cm.<br \/>\n\\( \\therefore \\frac{AP}{AB}=\\frac{3}{9}=\\frac{1}{3}\\text{ and }\\frac{AQ}{AC}=\\frac{5}{15}=\\frac{1}{3} \\)<br \/>\n\\( \\Rightarrow \\frac{AP}{AB}=\\frac{AQ}{AC} \\)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8604\/28955031692_17c69b57f5_o.png\" alt=\"similarity-of-triangles-17\" width=\"211\" height=\"146\" \/><br \/>\nThus, in triangles APQ and ABC, we have<br \/>\n\\( \\frac{AP}{AB}=\\frac{AQ}{AC} \\) \u00a0 \u00a0and \u2220A = \u2220A \u00a0 \u00a0 \u00a0 \u00a0[Common]<br \/>\nTherefore, by SAS-criterion of similarity, we have<br \/>\n\u2206APQ ~ \u2206ABC<br \/>\n\\( \\Rightarrow \\frac{AP}{AB}=\\frac{PQ}{BC}=\\frac{AQ}{AC} \\)<br \/>\n\\( \\Rightarrow \\frac{PQ}{BC}=\\frac{AQ}{AC}\\Rightarrow \\frac{PQ}{BC}=\\frac{5}{15} \\)<br \/>\n\\( \\Rightarrow \\frac{PQ}{BC}=\\frac{1}{3} \\)<br \/>\n\u21d2 BC = 3PQ<\/p>\n<p><strong>Example 12: \u00a0 \u00a0<\/strong>\u00a0In figure, \u2220A = \u2220CED, prove that \u2206CAB ~ \u2206CED. Also, find the value of x.<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8040\/28774133310_eaf0253a26_o.png\" alt=\"similarity-of-triangles-18\" width=\"234\" height=\"207\" \/><br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \u00a0In \u2206CAB and \u2206CED, we have<br \/>\n\u2220A = \u2220CED and \u2220C = \u2220C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [common]<br \/>\n\u2206CAB ~ \u2206CED<br \/>\n\\( \\Rightarrow \\frac{CA}{CE}=\\frac{AB}{DE}=\\frac{CB}{CD} \\)<br \/>\n\\( \\Rightarrow \\frac{AB}{DE}=\\frac{CB}{CD}\\Rightarrow \\frac{9}{x}=\\frac{10+2}{8} \\)<br \/>\n\u21d2 \u00a0x = 6 cm<\/p>\n<p><strong>Example 13: \u00a0 \u00a0<\/strong>\u00a0In the figure, E is a point on side CB produced of an isosceles \u2206ABC with AB = AC. If AD \u22a5\u00a0BC and EF \u22a5\u00a0AC, prove that \u2206ABD ~ \u2206ECF.<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8281\/28956939832_895184ff02_o.png\" alt=\"similarity-of-triangles-19\" width=\"174\" height=\"152\" \/><br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \u00a0<strong>Given:<\/strong> A \u2206ABC in which AB = AC and AD \u22a5\u00a0BC. Side CB is produced to E and EF \u22a5\u00a0AC.<br \/>\n<strong>To prove:<\/strong>\u00a0\u2206ABD ~ \u2206ECF.<br \/>\n<strong>Proof:<\/strong> we known that the angles opposite to equal sides of a triangle are equal.<br \/>\n\u2220B = \u2220C [\u2235\u00a0 AB = AC]<br \/>\nNow, in \u2206ABD and \u2206ECF, we have<br \/>\n\u2234 \u00a0\u2220B = \u2220C \u00a0 \u00a0 \u00a0 \u00a0[proved above]<br \/>\n\u2220ADB = \u2220EFC = 90\u00ba<br \/>\n\u2234 \u00a0\u2206ABD ~ \u2206ECF \u00a0 \u00a0 \u00a0 [By AA-similarity]<\/p>\n<p><strong>Example 14: \u00a0 \u00a0<\/strong>In figure, \u2220BAC = 90\u00ba and segment AD\u00a0\u22a5 BC. Prove that AD<sup>2<\/sup> = BD \u00d7 DC.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \u00a0In \u2206ABD and \u2206ACD, we have<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8332\/28444424683_2929675053_o.png\" alt=\"similarity-of-triangles-20\" width=\"221\" height=\"170\" \/><br \/>\n\u2220ADB = \u2220ADC \u00a0 \u00a0 \u00a0 [Each equal to 90\u00ba]<br \/>\nand, \u2220DBA = \u2220DAC<br \/>\n\\(\\left[ Each\\ equal\\ to\\ complement\\ of\\angle BAD\\ i.e.,\\ {{90}^{\\text{o}}}\\ -\\angle BAD \\right] \\)<br \/>\nTherefore, by AA-criterion of similarity, we have<br \/>\n\u2206DBA ~ \u2206DAC<br \/>\n\\(\\left[ \\therefore \\ \\ \\angle D\\leftrightarrow \\ \\angle D,\\ \\angle DBA\\leftrightarrow \\ \\angle DAC\\ and\\ \\angle BAD\\leftrightarrow \\ \\angle DCA \\right] \\)<br \/>\n\\(\\Rightarrow \\frac{DB}{DA}=\\frac{DA}{DC} \\)<br \/>\n\\(\\left[ In\\ similar\\ triangles\\ corresponding\\text{ }sides\\ are\\ proportional \\right] \\)<br \/>\n\\(\\Rightarrow \\frac{BD}{AD}=\\frac{AD}{DC}\u00a0\\)<br \/>\nAD<sup>2<\/sup>\u00a0= BD \u00d7 DC<\/p>\n<p><strong>Example 15: \u00a0 \u00a0<\/strong>In an isosceles \u2206ABC, the base AB is produced both ways in P and Q such that AP \u00d7 BQ = AC<sup>2<\/sup>\u00a0\u00a0and CE are the altitudes. Prove that \u2206ACP ~ \u2206BCQ.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> CA = CB \u00a0\u21d2\u00a0\u2220CAB = \u2220CBA<br \/>\n\u21d2 180\u00ba \u2013 \u2220CAB = 180\u00ba \u2013 \u2220CBA<br \/>\n\u21d2 \u2220CAP = \u2220CBQ<br \/>\nNow, AP \u00d7 BQ = AC<sup>2<\/sup><br \/>\n\\( \\Rightarrow \\frac{AP}{AC}=\\frac{AC}{BQ}\\Rightarrow \\frac{AP}{AC}=\\frac{BC}{BQ}\\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [\u2235\u00a0 AC = BC]<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8301\/28956939652_7aac7eef94_o.png\" alt=\"similarity-of-triangles-21\" width=\"171\" height=\"159\" \/><br \/>\nThus, \u2220CAP = \u2220CBQ \u00a0 \u00a0and \u00a0 \u00a0 \u00a0 \u00a0 \\( \\frac{AP}{AC}=\\frac{BC}{BQ} \\)<br \/>\n\u2234 \u00a0\u2206ACP ~ \u2206BCQ.<\/p>\n<p><strong>Example 16: \u00a0 \u00a0\u00a0<\/strong>The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F,\u00a0where E is any point on the side BC. Prove that DF \u00d7 EF = FB \u00d7 FA.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 In \u2206AFD and \u2206BFE, we have<br \/>\n\u22201 = \u22202 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Vertically opposite angles]<br \/>\n\u22203 = \u22204 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Alternate angles]<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c2.staticflickr.com\/8\/7594\/29029477456_1f0b9d81c9_o.png\" alt=\"similarity-of-triangles-22\" width=\"204\" height=\"147\" \/><br \/>\nSo, by AA-criterion of similarity, we have<br \/>\n\u2206FBE ~ \u2206FDA<br \/>\n\\( \\Rightarrow \\frac{FB}{FD}=\\frac{FD}{FA} \u00a0\\)<br \/>\n\\( \\Rightarrow \\frac{FB}{DF}=\\frac{EF}{FA} \\)<br \/>\n\u21d2 DF \u00d7 EF = FB \u00d7 FA<\/p>\n<p><strong>Example 17: \u00a0 \u00a0<\/strong>Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2 BL.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 In \u2206BMC and \u2206EMD, we have<br \/>\nMC = MD \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [\u2235 M is the mid-point of CD]<br \/>\n\u2220CMB = \u2220EMD \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Vertically opposite angles]<br \/>\nand, \u2220MBC = \u2220MED \u00a0 \u00a0 \u00a0 \u00a0 [Alternate angles]<br \/>\nSo, by AAS-criterion of congruence, we have<br \/>\n\u2234 \u2206BMC \u2245\u00a0\u2206EMD<br \/>\n\u21d2\u00a0BC = DE \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nAlso, AD = BC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\n[\u2235 ABCD is a parallelogram]<br \/>\nAD + DE = BC + BC<br \/>\n\u21d2\u00a0\u00a0AE = 2 BC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(iii)<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8316\/29062337575_b1518df363_o.png\" alt=\"similarity-of-triangles-23\" width=\"249\" height=\"130\" \/><br \/>\nNow, in \u2206AEL and \u2206CBL, we have<br \/>\n\u2220ALE = \u2220CLB \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Vertically opposite angles]<br \/>\n\u2220EAL = \u2220BCL \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Alternate angles]<br \/>\nSo, by AA-criterion of similarity of triangles,<br \/>\nwe have<br \/>\n\u2206AEL ~ \u2206CBL<br \/>\n\\( \\Rightarrow \\frac{EL}{BL}=\\frac{AE}{CB}\\Rightarrow \\frac{EL}{BL}=\\frac{2BC}{BC}\\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Using equations (iii)]<br \/>\n\\( \\Rightarrow \\frac{EL}{BL}=2 \\)<br \/>\n\u21d2\u00a0\u00a0EL = 2BL<\/p>\n<p><strong>Example 17: \u00a0 \u00a0\u00a0<\/strong>In figure, ABCD is a trapezium with AB || DC. If \u2206AED is similar to \u2206BEC, prove that AD = BC.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 In \u2206EDC and \u2206EBA, we have<br \/>\n\u22201 = \u22202 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Alternate angles]<br \/>\n\u22203 = \u22204 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Alternate angles]<br \/>\nand, \u2220CED =\u00a0\u2220AEB \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Vertically opposite angles]<br \/>\n\u2234 \u00a0\u2206EDC ~ \u2206EBA<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8208\/28956939432_91c07ce751_o.png\" alt=\"similarity-of-triangles-24\" width=\"160\" height=\"153\" \/><br \/>\n\\( \\Rightarrow \\frac{ED}{EB}=\\frac{EC}{EA} \\)<br \/>\n\\( \\Rightarrow \\frac{ED}{EB}=\\frac{EB}{EA} \\)\u00a0 \u00a0 \u00a0&#8230;.(i)<br \/>\nIt is given that \u2206AED ~ \u2206BEC<br \/>\n\\( \\Rightarrow \\frac{ED}{EB}=\\frac{EA}{EB}=\\frac{AD}{BC} \\) \u00a0 \u00a0 \u00a0&#8230;.(ii)<br \/>\nFrom (i) and (ii), we get<br \/>\n\\( \\frac{EB}{EA}=\\frac{EA}{EB} \\)<br \/>\n\u21d2\u00a0\u00a0\u00a0(EB)<sup>2<\/sup> = (EA)<sup>2<\/sup><br \/>\n\u21d2\u00a0\u00a0\u00a0EB = EA<br \/>\nSubstituting EB = EA in (ii), we get<br \/>\n\\(\\frac{EA}{EA}=\\frac{AD}{BC}\\Rightarrow \\frac{AD}{BC}=1\\)<br \/>\n\u21d2 \u00a0 AD = BC<\/p>\n<p><strong>Example 18: \u00a0 \u00a0\u00a0<\/strong>A vertical stick 20 cm long casts a shadow 6 cm long on the ground. At the same time, a tower casts a shadow 15 m long on the ground. Find the height of the tower.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let the sun\u2019s altitude at that moment be \u03b8.<br \/>\n\u2206PQM ~ \u2206ABC<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8125\/29029477446_42e42e7c7c_o.png\" alt=\"similarity-of-triangles-25\" width=\"333\" height=\"253\" \/><br \/>\n\\( \\Rightarrow \\frac{MP}{MQ}=\\frac{AC}{CB} \\)<br \/>\n\\( \\Rightarrow \\frac{h}{15}=\\frac{20}{6} \\)<br \/>\n\u2234 \u00a0Height of the tower = 50 m.<\/p>\n<p><strong>Example 19: \u00a0 \u00a0\u00a0<\/strong>If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar to each other and to the original triangle. Also, prove that the square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 <strong>Given:<\/strong> A right triangle ABC right angled at B, BD \u22a5\u00a0AC.<br \/>\n<strong>To Prove:<\/strong><br \/>\n(i) \u2206ADB ~ \u2206BDC \u00a0 \u00a0 \u00a0 (ii) \u2206ADB ~ \u2206ABC<br \/>\n(iii) \u2206BDC ~ \u2206ABC \u00a0 \u00a0(iv) BD<sup>2<\/sup> = AD \u00d7 DC<br \/>\n(v) AB<sup>2<\/sup>= AD \u00d7 AC \u00a0 \u00a0 \u00a0(vi) BC<sup>2<\/sup> = CD \u00d7 AC<br \/>\n<strong>Proof:<\/strong><br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8059\/28956939192_bf257626b0_o.png\" alt=\"similarity-of-triangles-26\" width=\"209\" height=\"156\" \/><br \/>\n<strong>(i) \u00a0<\/strong>We have,<br \/>\n\u2220ABD + \u2220DBC = 90\u00ba<br \/>\nAlso, \u2220C + \u2220DBC + \u2220BDC = 180\u00ba<br \/>\n\u21d2\u00a0\u00a0\u00a0\u2220C + \u2220DBC + 90\u00ba = 180\u00ba<br \/>\n\u21d2\u00a0\u00a0\u00a0\u2220C + \u2220DBC = 90\u00ba<br \/>\nBut, \u2220ABD + \u2220DBC = 90\u00ba<br \/>\n\u2234 \u2220ABD + \u2220DBC = \u2220C + \u2220DBC<br \/>\n\u21d2\u00a0\u00a0\u2220ABD = \u2220C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(i)<br \/>\nThus, in \u2206ADB and \u2206BDC, we have<br \/>\n\u2220ABD = \u2220C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [From (i)]<br \/>\nand, \u2220ADB = \u2220BDC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Each equal to 90\u00ba]<br \/>\nSo, by AA-similarity criterion, we have<br \/>\n\u2206ADB ~ \u2206BDC<br \/>\n<strong>(ii) \u00a0<\/strong>In \u2206ADB and \u2206ABC, we have<br \/>\n\u2220ADB = \u2220ABC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Each equal to 90\u00ba]<br \/>\nand, \u2220A = \u2220A \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Common]<br \/>\nSo, by AA-similarity criterion, we have<br \/>\n\u2206ADB ~ \u2206ABC<br \/>\n<strong>(iii) \u00a0<\/strong>In \u2206BDC and \u2206ABC, we have<br \/>\n\u2220BDC = \u2220ABC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Each equal to 90\u00ba]<br \/>\n\u2220C = \u2220C \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Common]<br \/>\nSo, by AA-similarity criterion, we have<br \/>\n\u2206BDC ~ \u2206ABC<br \/>\n<strong>(iv) \u00a0<\/strong>From (i), we have<br \/>\n\u2206ADB ~ \u2206BDC<br \/>\n\\( \\Rightarrow \\frac{AD}{BD}=\\frac{BD}{DC}\\)<br \/>\n\u21d2 BD<sup>2<\/sup> = AD \u00d7 DC<br \/>\n<strong>(v) \u00a0<\/strong>From (ii), we have<br \/>\n\\( \\Rightarrow \\frac{AD}{AB}=\\frac{AB}{AC}\\)<br \/>\n\u2206ADB ~ \u2206ABC<br \/>\n\u21d2 AB<sup>2<\/sup> = AD \u00d7 AC<br \/>\n<strong>(vi) \u00a0<\/strong>From (iii), we have<br \/>\n\\( \\Rightarrow \\frac{BC}{AC}=\\frac{DC}{BC}\\)<br \/>\n\u2206BDC ~ \u2206ABC<br \/>\n\u21d2 BC<sup>2<\/sup> = CD \u00d7 AC<\/p>\n<p><strong>Example 20: \u00a0 \u00a0\u00a0<\/strong>Prove that the line segments joining the mid points of the sides of a triangle form four triangles, each of which is similar to the original triangle.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \u00a0<strong>Given:<\/strong>\u00a0\u2206ABC in which D, E, F are the<br \/>\nmid-points of sides BC, CA and AB respectively.<br \/>\n<strong>To Prove:<\/strong> Each of the triangles AFE, FBD, EDC and DEF is similar to \u2206ABC.<br \/>\n<strong>Proof:<\/strong> Consider triangles AFE and ABC.<br \/>\nSince F and E are mid-points of AB and AC respectively.<br \/>\n\u2234 \u00a0FE || BC<br \/>\n\u21d2 \u2220AEF = \u2220B \u00a0 \u00a0 \u00a0\u00a0[Corresponding angles]<br \/>\nThus, in \u2206AFE and \u2206ABC, we have<br \/>\n\u2220AFE = \u2220B<br \/>\nand, \u2220A = \u2220A \u00a0 \u00a0 \u00a0 \u00a0[Common]<br \/>\n\u2234 \u00a0\u2206AFE ~ \u2206ABC.<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8366\/29062337365_0557081c7a_o.png\" alt=\"similarity-of-triangles-27\" width=\"154\" height=\"160\" \/><br \/>\nSimilarly, we have<br \/>\n\u2206FBD ~ \u2206ABC and \u2206EDC ~ \u2206ABC.<br \/>\nNow, we shall show that \u2206DEF ~ \u2206ABC.<br \/>\nClearly, ED || AF and DE || EA.<br \/>\n\u2234 \u00a0\u00a0AFDE is a parallelogram.<br \/>\n\u21d2 \u00a0 \u2220EDF = \u2220A<br \/>\n[\u2235\u00a0 Opposite angles of a parallelogram are equal]<br \/>\nSimilarly, BDEF is a parallelogram.<br \/>\n\u2234 \u00a0\u2220DEF = \u2220B<br \/>\n[\u2235\u00a0 Opposite angles of a parallelogram are equal]<br \/>\nThus, in triangles DEF and ABC, we have<br \/>\n\u2220EDF = \u2220A and \u2220DEF = \u2220B<br \/>\nSo, by AA-criterion of similarity, we have<br \/>\n\u2206DEF ~ \u2206ABC.<br \/>\nThus, each one of the triangles AFE, FBD, EDC and DEF is similar to \u2206ABC.<\/p>\n<p><strong>Example 21: \u00a0 \u00a0\u00a0<\/strong>In \u2206ABC, DE is parallel to base BC, with D on AB and E on AC. If \u00a0 \\(\\frac{AD}{DB}=\\frac{2}{3}\\) , find \\(\\frac{BC}{DE}\\).<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \u00a0In \u2206ABC, we have<br \/>\n<img loading=\"lazy\" src=\"https:\/\/c1.staticflickr.com\/9\/8376\/28956939022_b0da82635c_o.png\" alt=\"similarity-of-triangles-28\" width=\"161\" height=\"159\" \/><br \/>\nDE || BC \u00a0 \u00a0\u00a0\\(\\Rightarrow \\frac{AB}{AD}=\\frac{AC}{AE}\\)<br \/>\nThus, in triangles ABC and ADE, we have<br \/>\n\\(\\frac{AB}{AD}=\\frac{AC}{AE}\\) \u00a0 \u00a0 \u00a0and, \u2220A = \u2220A<br \/>\nTherefore, by SAS-criterion of similarity, we have<br \/>\n\u2206ABC ~ \u2206ADE<br \/>\n\\(\\Rightarrow \\frac{AD}{AD}=\\frac{BC}{DE}\\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nIt is given that<br \/>\n\\( \\frac{AD}{DB}=\\frac{2}{3} \u00a0\\)<br \/>\n\\( \\Rightarrow \\frac{DB}{AD}=\\frac{3}{2} \\)<br \/>\n\\( \\Rightarrow \\frac{DB}{AD}+1=\\frac{3}{2}+1 \\)<br \/>\n\\( \\Rightarrow \\frac{DB+AD}{AD}=\\frac{5}{2} \\)<br \/>\n\\( \\Rightarrow \\frac{AB}{DE}=\\frac{5}{2} \u00a0\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(ii)<br \/>\nFrom (i) and (ii), we get<br \/>\n\\( \\frac{BC}{DE}=\\frac{5}{2} \\)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Criteria For Similarity Of Triangles AAA similarity criterion: If in two triangles, corresponding angles are equal, then the triangles are similar. AA Similarity criterion: If in two triangles, two angles of one triangle are respectively equal the two angles of the other triangle, then the two triangles are similar. SSS Similarity criterion: If in two &#8230; <a title=\"Criteria For Similarity Of Triangles\" class=\"read-more\" href=\"https:\/\/cbselibrary.com\/similarity-of-triangles\/\" aria-label=\"More on Criteria For Similarity Of Triangles\">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":[76,77,71],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v18.4.1) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Criteria For Similarity Of Triangles - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/similarity-of-triangles\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Criteria For Similarity Of Triangles\" \/>\n<meta property=\"og:description\" content=\"Criteria For Similarity Of Triangles AAA similarity criterion: If in two triangles, corresponding angles are equal, then the triangles are similar. AA Similarity criterion: If in two triangles, two angles of one triangle are respectively equal the two angles of the other triangle, then the two triangles are similar. SSS Similarity criterion: If in two ... 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