Question

In the given figure, AD = AE and AD² = BD × EC. Prove that: triangles ABD and CAE are similar. 

Answer 1

In the given figure,

AD = AE

AD² = BD × EC

To prove: ΔABD ~ ΔCAE

Proof: In ΔADC, AD = AE

∴ ∠ADE = ∠AED  ...(Angles opposite to equal sides)

 But ∠ADE + ∠ADB

= ∠AED + ∠AEC

= 180°

∴ ∠ADB = ∠AEC

AD² = BD × EC

`(AD)/(BD) = (EC)/(AD)`

`\implies (AE)/(BD) = (EC)/(AD)`  ...(∵ AD = AE)

And ∠ADB = ∠AEC

∴ ΔABD ~ ΔCAE   ...(SAS axiom)