AA Similarity
AA similarity : If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.Paragraph proof :
Let ΔABC and ΔDEF be two triangles such that ∠A = ∠D and ∠B = ∠E.
∠A + ∠B + ∠C = 180 ^{0} (Sum of all angles in a Δ is 180)
∠D + ∠E + ∠F = 180 ^{0} (Sum of all angles in a Δ is 180)
⇒ ∠A + ∠B + ∠C = ∠D + ∠E + ∠F
⇒ ∠D + ∠E + ∠C = ∠D + ∠E + ∠F (since ∠A = ∠D and ∠B = ∠E)
⇒ ∠C = ∠F
Thus the two triangles are equiangular and hence they are similar by AA.
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Examples
1) D is a point on the side of BC of ΔABC such that ∠ADC = ∠BAC.
Prove that CA ^{2} = BA x CD
Given : ∠ADC = ∠BAC
Prove that : CA ^{2} = BA x CD


1) ∠ADC = ∠BAC  1) Given 
2) ∠C = ∠C  2) Reflexive (common) 
3) ΔABC ~ ΔDAC  3) AA criteria (postulate) 
4) AB/DA = CB/CA = CA/CD  4) If two triangles are similar then their sides are in proportion. 
5) CB/CA = CA /CD  5) Last two ratios 
6) CA^{2} = CB x CD  6) Cross multiplication . 
2) In the given figure, DEBC such that AE=(1/4)AC. If AB= 6 cm, then find the value of AD.
Solution : As DE  BC,
AE=(1/4)AC
⇒ AE/AC = 1/4
ΔABC ~ ΔADE > AAsimilarity
∴ AD/AB = AE/AC
AD/6 = 1/4 AD = (6 x1)/4
AD = 1.5cm.
Criteria for Similarity
• AAA Similarity
• AA Similarity
• SSS Similarity
• SAS Similarity
• Practice on Similarity
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