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03/06/2021 2:40 pm
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If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.
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03/06/2021 2:42 pm
Given, ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.
∴ AB/PQ = AC/PR = BC/QR ……………(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii)
Since AD and PM are medians, they will divide their opposite sides.
∴ BD = BC/2 and QM = QR/2 ……………..………….(iii)
From equations (i) and (iii)
AB/PQ = BD/QM ……………………….(iv)
In ΔABD and ΔPQM,
From equation (ii)
∠B = ∠Q
From equation (iv),
AB/PQ = BD/QM
∴ ΔABD ~ ΔPQM (SAS similarity criterion)
⇒AB/PQ = BD/QM = AD/PM
This post was modified 4 years ago by Raavi Tiwari