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