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Question
Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that : `(AB)/(PQ) = (AD)/(PM)`.
Solution
Given, ΔABC ∼ ΔPQR
AD and PM are the angle bisectors of ∠A and ∠P respectively.
To prove: `(AB)/(PQ) = (AD)/(PM)`
Proof: Since, ΔABC ∼ ΔPQR
∴ ∠A = ∠P and ∠B = ∠Q
`(AB)/(PQ) = (BC)/(QR)`
∵ ∠A = ∠P
`\implies 1/2 ∠A = 1/2 ∠P`
`\implies` ∠BAD = ∠QPM
Now in ΔABD and ΔPQM
∠B = ∠Q
∠BAD = ∠QPM ...(Proved)
∴ ΔBAD ∼ ΔQPM ...(AA axiom)
∴ `(AB)/(PQ) = (AD)/(PM)`
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