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 triangles, corresponding sides are in the same ratio, then the two triangles are similar.
SAS similarity criterion: If 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.
Equiangular Triangles:
Two triangles are said to be equiangular, if their corresponding angles are equal.
If two triangles are equiangular, then they are similar.
Two triangles ABC and DEF such that
∠A = ∠D, ∠B = ∠E and ∠C = ∠F.
Then ∆ABC ~ ∆DEF and
\(\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\)
Read More:
- Angle Sum Property of a Triangle
- Median and Altitude of a Triangle
- The Angle of An Isosceles Triangle
- Areas of Two Similar Triangles
- Area of A Triangle
- To Prove Triangles Are Congruent
- Construction of an Equilateral Triangle
- Classification of Triangles
Criteria For Similarity Of Triangles With Examples
Example 1: In figure, find ∠L.
Sol. In ∆ABC and ∆LMN,
\( \frac{AB}{LM}=\frac{4.4}{11}=\frac{2}{5}\)
\( \frac{BC}{MN}=\frac{4}{10}=\frac{2}{5} \)
\( \frac{CA}{NL}=\frac{3.6}{9}=\frac{2}{5} \)
\( \Rightarrow \frac{AB}{LM}=\frac{BC}{MN}=\frac{CA}{NL} \)
∆ABC ~ ∆LMN (SSS similarity)
∠L = ∠A
= 180º – ∠B – ∠C
= 180º – 50º – 70º = 60º
∠L = 60º
Example 2: Examine each pair of triangles in Figure, and state which pair of triangles are similar. Also, state the similarity criterion used by you for answering the question and write the similarity relation in symbolic form.
Figure (i)
Figure (ii)
Figure (iii)
Figure (iv)
Figure (v)
Figure (vi)
Figure (vii)
Sol. (i) ∠A = ∠Q, ∠B = ∠P and ∠C = ∠R.
∴ ∆ABC ~ ∆QPR (AAA-similarity)
(ii) In triangle PQR and DEF, we observe that
\(\frac{PQ}{DE}=\frac{QR}{EF}=\frac{PR}{DF}=\frac{1}{2} \)
Therefore, by SSS-criterion of similarity, we have
∆PQR ~ ∆DEF
(iii) SAS-similarity is not satisfied as included angles are not equal.
(iv) ∆CAB ~ ∆QRP (SAS-similarity), as
\(\frac{CA}{QR}=\frac{CB}{QP} \) and ∠C = ∠Q.
(v) In ∆’s ABC and DEF, we have
∠A = ∠D = 80º
\(\text{But, }\frac{AB}{DE}\ne \frac{AC}{DF} \) [∵ AC is not given]
So, by SAS-criterion of similarity these two triangles are not similar.
(vi) In ∆’s DEF and MNP, we have
∠D = ∠M = 70º
∠E = ∠N = 80º [∵ ∠N = 180º – ∠M – ∠P = 180º – 70º – 30º = 80º]
So, by AA-criterion of similarity
∆DEF ~ ∆MNP.
(vii) FE = 2 cm, FD = 3 cm, ED = 2.5 cm
PQ = 4 cm, PR = 6 cm, QR = 5 cm
∴ ∆FED ~ ∆PQR (SSS-similarity)
Example 3: In figure, QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Find AQ.
Sol. In triangles AOQ and BOP, we have
∠OAQ = ∠OBP [Each equal to 90º]
∠AOQ = ∠BOP [Vertically opposite angles]
Therefore, by AA-criterion of similarity
∆AOQ ~ ∆BOP
\( \Rightarrow \frac{AO}{BO}=\frac{OQ}{OP}=\frac{AO}{BP} \)
\( \Rightarrow \frac{AO}{BO}=\frac{AQ}{BP}\Rightarrow \frac{10}{6}=\frac{AQ}{9} \)
\( \Rightarrow AQ=\frac{10\times 9}{6}=15\text{ }cm \)
Example 4: In figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, find CA and AQ.
Sol. We have, ∆ACB ~ ∆APQ
\( \Rightarrow \frac{AC}{AP}=\frac{CB}{PQ}=\frac{AB}{AQ} \)
\( \Rightarrow \frac{AC}{AP}=\frac{CB}{PQ}\text{ and }\frac{CB}{PQ}=\frac{AB}{AQ} \)
\( \Rightarrow \frac{AC}{2.8}=\frac{8}{4}\text{ and }\frac{8}{4}=\frac{6.5}{AQ} \)
\( \Rightarrow \frac{AC}{2.8}=2\text{ and }\frac{6.5}{AQ}=2 \)
AC = (2 × 2.8) cm = 5.6 cm and \( AQ=\frac{6.5}{2}\text{ }cm\text{ }=\text{ }3.25\text{ }cm \)
Example 5: The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find AB.
Sol. Since the ratio of the corresponding sides of similar triangles is same as the ratio of their perimeters.
∴ ∆ABC ~ ∆PQR
\( \Rightarrow \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{36}{24} \)
\( \Rightarrow \frac{AB}{PQ}=\frac{36}{24}\Rightarrow \frac{AB}{10}=\frac{36}{24} \)
\( \Rightarrow AB=\frac{36\times 10}{24}cm\text{ }=\text{ }15\text{ }cm \)
Example 6: In figure, ∠CAB = 90º and AD ⊥ BC. If AC = 75 cm, AB = 1 m and BD = 1.25 m, find AD.
Sol. We have,
AB = 1 m = 100 cm, AC = 75 cm and BD = 125 cm
In ∆BAC and ∆BDA, we have
∠BAC = ∠BDA [Each equal to 90º]
and, ∠B = ∠B
So, by AA-criterion of similarity, we have
∆BAC ~ ∆BDA
\( \Rightarrow \frac{BA}{BD}=\frac{AC}{AD} \)
\( \Rightarrow \frac{100}{125}=\frac{75}{AD} \)
\( \Rightarrow AD=\frac{125\times 75}{100}cm\text{ }=\text{ }93.75\text{ }cm \)
Example 7: In figure, if ∠A = ∠C, then prove that ∆AOB ~ ∆COD.
Sol. In triangles AOB and COD, we have
∠A = ∠C [Given]
and, ∠1 = ∠2 [Vertically opposite angles]
Therefore, by AA-criterion of similarity, we have
∆AOB ~ ∆COD
Example 8: In figure, \( \frac{AO}{OC}=\frac{BO}{OD}=\frac{1}{2} \) and AB = 5 cm. Find the value of DC.
Sol. In ∆AOB and ∆COD, we have
∆AOB = ∆COD [Vertically opposite angles]
\( \frac{AO}{OC}=\frac{BO}{OD} \) [Given]
So, by SAS-criterion of similarity, we have
∆AOB ~ ∆COD
\( \Rightarrow \frac{AO}{OC}=\frac{BO}{OD}=\frac{AB}{DC} \)
\( \Rightarrow \frac{1}{2}=\frac{5}{DC} \) [∵ AB = 5 cm]
⇒ DC = 10 cm
Example 9: In figure, considering triangles BEP and CPD, prove that BP × PD = EP × PC.
Sol. Given: A ∆ABC in which BD ⊥ AC and CE ⊥ AB and BD and CE intersect at P.
To Prove: BP × PD = EP × PC
Proof: In ∆EPB and ∆DPC, we have
∠PEB = ∠PDC [Each equal to 90º]
∠EPB = ∠DPC [Vertically opposite angles]
Thus, by AA-criterion of similarity, we have
∆EPB ~ ∆DPC
\(\frac{EP}{DP}=\frac{PB}{PC}\)
⇒ BP × PD = EP × PC
Example 10: D is a point on the side BC of ∆ABC such that ∠ADC = ∠BAC.
Prove that \(\frac{CA}{CD}=\frac{CB}{CA}\) or, CA2 = CB × CD.
Sol. In ∆ABC and ∆DAC, we have
∠ADC = ∠BAC and ∠C = ∠C
Therefore, by AA-criterion of similarity, we have
∆ABC ~ ∆DAC
\( \Rightarrow \frac{AB}{DA}=\frac{BC}{AC}=\frac{AC}{DC} \)
\( \Rightarrow \frac{CB}{CA}=\frac{CA}{CD} \)
Example 11: P and Q are points on sides AB and AC respectively of ∆ABC. If AP = 3 cm, PB = 6cm. AQ = 5 cm and QC = 10 cm, show that BC = 3PQ.
Sol. We have,
AB = AP + PB = (3 + 6) cm = 9 cm
and, AC = AQ + QC = (5 + 10) cm = 15 cm.
\( \therefore \frac{AP}{AB}=\frac{3}{9}=\frac{1}{3}\text{ and }\frac{AQ}{AC}=\frac{5}{15}=\frac{1}{3} \)
\( \Rightarrow \frac{AP}{AB}=\frac{AQ}{AC} \)
Thus, in triangles APQ and ABC, we have
\( \frac{AP}{AB}=\frac{AQ}{AC} \) and ∠A = ∠A [Common]
Therefore, by SAS-criterion of similarity, we have
∆APQ ~ ∆ABC
\( \Rightarrow \frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC} \)
\( \Rightarrow \frac{PQ}{BC}=\frac{AQ}{AC}\Rightarrow \frac{PQ}{BC}=\frac{5}{15} \)
\( \Rightarrow \frac{PQ}{BC}=\frac{1}{3} \)
⇒ BC = 3PQ
Example 12: In figure, ∠A = ∠CED, prove that ∆CAB ~ ∆CED. Also, find the value of x.
Sol. In ∆CAB and ∆CED, we have
∠A = ∠CED and ∠C = ∠C [common]
∆CAB ~ ∆CED
\( \Rightarrow \frac{CA}{CE}=\frac{AB}{DE}=\frac{CB}{CD} \)
\( \Rightarrow \frac{AB}{DE}=\frac{CB}{CD}\Rightarrow \frac{9}{x}=\frac{10+2}{8} \)
⇒ x = 6 cm
Example 13: In the figure, E is a point on side CB produced of an isosceles ∆ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.
Sol. Given: A ∆ABC in which AB = AC and AD ⊥ BC. Side CB is produced to E and EF ⊥ AC.
To prove: ∆ABD ~ ∆ECF.
Proof: we known that the angles opposite to equal sides of a triangle are equal.
∠B = ∠C [∵ AB = AC]
Now, in ∆ABD and ∆ECF, we have
∴ ∠B = ∠C [proved above]
∠ADB = ∠EFC = 90º
∴ ∆ABD ~ ∆ECF [By AA-similarity]
Example 14: In figure, ∠BAC = 90º and segment AD ⊥ BC. Prove that AD2 = BD × DC.
Sol. In ∆ABD and ∆ACD, we have
∠ADB = ∠ADC [Each equal to 90º]
and, ∠DBA = ∠DAC
\(\left[ Each\ equal\ to\ complement\ of\angle BAD\ i.e.,\ {{90}^{\text{o}}}\ -\angle BAD \right] \)
Therefore, by AA-criterion of similarity, we have
∆DBA ~ ∆DAC
\(\left[ \therefore \ \ \angle D\leftrightarrow \ \angle D,\ \angle DBA\leftrightarrow \ \angle DAC\ and\ \angle BAD\leftrightarrow \ \angle DCA \right] \)
\(\Rightarrow \frac{DB}{DA}=\frac{DA}{DC} \)
\(\left[ In\ similar\ triangles\ corresponding\text{ }sides\ are\ proportional \right] \)
\(\Rightarrow \frac{BD}{AD}=\frac{AD}{DC} \)
AD2 = BD × DC
Example 15: In an isosceles ∆ABC, the base AB is produced both ways in P and Q such that AP × BQ = AC2 and CE are the altitudes. Prove that ∆ACP ~ ∆BCQ.
Sol. CA = CB ⇒ ∠CAB = ∠CBA
⇒ 180º – ∠CAB = 180º – ∠CBA
⇒ ∠CAP = ∠CBQ
Now, AP × BQ = AC2
\( \Rightarrow \frac{AP}{AC}=\frac{AC}{BQ}\Rightarrow \frac{AP}{AC}=\frac{BC}{BQ}\) [∵ AC = BC]
Thus, ∠CAP = ∠CBQ and \( \frac{AP}{AC}=\frac{BC}{BQ} \)
∴ ∆ACP ~ ∆BCQ.
Example 16: The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Prove that DF × EF = FB × FA.
Sol. In ∆AFD and ∆BFE, we have
∠1 = ∠2 [Vertically opposite angles]
∠3 = ∠4 [Alternate angles]
So, by AA-criterion of similarity, we have
∆FBE ~ ∆FDA
\( \Rightarrow \frac{FB}{FD}=\frac{FD}{FA} \)
\( \Rightarrow \frac{FB}{DF}=\frac{EF}{FA} \)
⇒ DF × EF = FB × FA
Example 17: 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.
Sol. In ∆BMC and ∆EMD, we have
MC = MD [∵ M is the mid-point of CD]
∠CMB = ∠EMD [Vertically opposite angles]
and, ∠MBC = ∠MED [Alternate angles]
So, by AAS-criterion of congruence, we have
∴ ∆BMC ≅ ∆EMD
⇒ BC = DE ….(i)
Also, AD = BC ….(ii)
[∵ ABCD is a parallelogram]
AD + DE = BC + BC
⇒ AE = 2 BC ….(iii)
Now, in ∆AEL and ∆CBL, we have
∠ALE = ∠CLB [Vertically opposite angles]
∠EAL = ∠BCL [Alternate angles]
So, by AA-criterion of similarity of triangles,
we have
∆AEL ~ ∆CBL
\( \Rightarrow \frac{EL}{BL}=\frac{AE}{CB}\Rightarrow \frac{EL}{BL}=\frac{2BC}{BC}\) [Using equations (iii)]
\( \Rightarrow \frac{EL}{BL}=2 \)
⇒ EL = 2BL
Example 17: In figure, ABCD is a trapezium with AB || DC. If ∆AED is similar to ∆BEC, prove that AD = BC.
Sol. In ∆EDC and ∆EBA, we have
∠1 = ∠2 [Alternate angles]
∠3 = ∠4 [Alternate angles]
and, ∠CED = ∠AEB [Vertically opposite angles]
∴ ∆EDC ~ ∆EBA
\( \Rightarrow \frac{ED}{EB}=\frac{EC}{EA} \)
\( \Rightarrow \frac{ED}{EB}=\frac{EB}{EA} \) ….(i)
It is given that ∆AED ~ ∆BEC
\( \Rightarrow \frac{ED}{EB}=\frac{EA}{EB}=\frac{AD}{BC} \) ….(ii)
From (i) and (ii), we get
\( \frac{EB}{EA}=\frac{EA}{EB} \)
⇒ (EB)2 = (EA)2
⇒ EB = EA
Substituting EB = EA in (ii), we get
\(\frac{EA}{EA}=\frac{AD}{BC}\Rightarrow \frac{AD}{BC}=1\)
⇒ AD = BC
Example 18: 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.
Sol. Let the sun’s altitude at that moment be θ.
∆PQM ~ ∆ABC
\( \Rightarrow \frac{MP}{MQ}=\frac{AC}{CB} \)
\( \Rightarrow \frac{h}{15}=\frac{20}{6} \)
∴ Height of the tower = 50 m.
Example 19: 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.
Sol. Given: A right triangle ABC right angled at B, BD ⊥ AC.
To Prove:
(i) ∆ADB ~ ∆BDC (ii) ∆ADB ~ ∆ABC
(iii) ∆BDC ~ ∆ABC (iv) BD2 = AD × DC
(v) AB2= AD × AC (vi) BC2 = CD × AC
Proof:
(i) We have,
∠ABD + ∠DBC = 90º
Also, ∠C + ∠DBC + ∠BDC = 180º
⇒ ∠C + ∠DBC + 90º = 180º
⇒ ∠C + ∠DBC = 90º
But, ∠ABD + ∠DBC = 90º
∴ ∠ABD + ∠DBC = ∠C + ∠DBC
⇒ ∠ABD = ∠C ….(i)
Thus, in ∆ADB and ∆BDC, we have
∠ABD = ∠C [From (i)]
and, ∠ADB = ∠BDC [Each equal to 90º]
So, by AA-similarity criterion, we have
∆ADB ~ ∆BDC
(ii) In ∆ADB and ∆ABC, we have
∠ADB = ∠ABC [Each equal to 90º]
and, ∠A = ∠A [Common]
So, by AA-similarity criterion, we have
∆ADB ~ ∆ABC
(iii) In ∆BDC and ∆ABC, we have
∠BDC = ∠ABC [Each equal to 90º]
∠C = ∠C [Common]
So, by AA-similarity criterion, we have
∆BDC ~ ∆ABC
(iv) From (i), we have
∆ADB ~ ∆BDC
\( \Rightarrow \frac{AD}{BD}=\frac{BD}{DC}\)
⇒ BD2 = AD × DC
(v) From (ii), we have
\( \Rightarrow \frac{AD}{AB}=\frac{AB}{AC}\)
∆ADB ~ ∆ABC
⇒ AB2 = AD × AC
(vi) From (iii), we have
\( \Rightarrow \frac{BC}{AC}=\frac{DC}{BC}\)
∆BDC ~ ∆ABC
⇒ BC2 = CD × AC
Example 20: 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.
Sol. Given: ∆ABC in which D, E, F are the
mid-points of sides BC, CA and AB respectively.
To Prove: Each of the triangles AFE, FBD, EDC and DEF is similar to ∆ABC.
Proof: Consider triangles AFE and ABC.
Since F and E are mid-points of AB and AC respectively.
∴ FE || BC
⇒ ∠AEF = ∠B [Corresponding angles]
Thus, in ∆AFE and ∆ABC, we have
∠AFE = ∠B
and, ∠A = ∠A [Common]
∴ ∆AFE ~ ∆ABC.
Similarly, we have
∆FBD ~ ∆ABC and ∆EDC ~ ∆ABC.
Now, we shall show that ∆DEF ~ ∆ABC.
Clearly, ED || AF and DE || EA.
∴ AFDE is a parallelogram.
⇒ ∠EDF = ∠A
[∵ Opposite angles of a parallelogram are equal]
Similarly, BDEF is a parallelogram.
∴ ∠DEF = ∠B
[∵ Opposite angles of a parallelogram are equal]
Thus, in triangles DEF and ABC, we have
∠EDF = ∠A and ∠DEF = ∠B
So, by AA-criterion of similarity, we have
∆DEF ~ ∆ABC.
Thus, each one of the triangles AFE, FBD, EDC and DEF is similar to ∆ABC.
Example 21: In ∆ABC, DE is parallel to base BC, with D on AB and E on AC. If \(\frac{AD}{DB}=\frac{2}{3}\) , find \(\frac{BC}{DE}\).
Sol. In ∆ABC, we have
DE || BC \(\Rightarrow \frac{AB}{AD}=\frac{AC}{AE}\)
Thus, in triangles ABC and ADE, we have
\(\frac{AB}{AD}=\frac{AC}{AE}\) and, ∠A = ∠A
Therefore, by SAS-criterion of similarity, we have
∆ABC ~ ∆ADE
\(\Rightarrow \frac{AD}{AD}=\frac{BC}{DE}\) ….(i)
It is given that
\( \frac{AD}{DB}=\frac{2}{3} \)
\( \Rightarrow \frac{DB}{AD}=\frac{3}{2} \)
\( \Rightarrow \frac{DB}{AD}+1=\frac{3}{2}+1 \)
\( \Rightarrow \frac{DB+AD}{AD}=\frac{5}{2} \)
\( \Rightarrow \frac{AB}{DE}=\frac{5}{2} \) ….(ii)
From (i) and (ii), we get
\( \frac{BC}{DE}=\frac{5}{2} \)