# 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 = 1800 (Sum of all angles in a Δ is 180)

∠D + ∠E + ∠F = 1800(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 CA2 = BA x CD
Prove that : CA2 = BA x CD Statements Reasons 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) CA2 = CB x CD 6) Cross multiplication .
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2) In the given figure, DE||BC 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