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Question
Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that : `(AB)/(PQ) = (AD)/(PM)`.
Solution
Given, ΔABC ∼ ΔPQR
AD and PM are altitudes of these two triangles
To prove: `(AB)/(PQ) = (AD)/(PM)`
Proof: Since, ΔABC ∼ ΔPQR
∴ ∠B = ∠Q
`(AB)/(PQ) = (BC)/(QR)`
Now in ΔABD and ΔPQM
∠B = ∠Q
∠D = ∠M ...(Each 90°)
∴ ΔABD ∼ ΔPQM ...(AAS axiom)
∴ `(AB)/(PQ) = (AD)/(PM)` ...(Corresponding sides of Δ's are proportional)
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