If AD and PM are medians of triangles ABC and PQR, respectively where \(\triangle\) ABC ~ \(\triangle\) PQR prove that AB/PQ = AD/PM.

Given, \(\triangle\) ABC ~ \(\triangle\) PQR



We know that the corresponding sides of similar triangles are in proportion.

AB/PQ = AC/PR = BC/QR……………………………(i)
Also, \(\angle\) A = \(\angle\) P, \(\angle\) B = \(\angle\) Q, \(\angle\) C = \(\angle\) R ………….…..(ii)

Since AD and PM are medians, they will divide their opposite sides.
BD = \(\frac{BC}{2}\) and QM = \(\frac{QR}{2}\) ……………..………….(iii)

From equations (i) and (iii), we get
AB/PQ = BD/QM ……………………….(iv)

In \(\triangle\) ABD and \(\triangle\) PQM,

From equation (ii), we have
\(\angle\) B = \(\angle\) Q

From equation (iv), we have,
AB/PQ = BD/QM
\(\triangle\) ABD ~ \(\triangle\) PQM (SAS similarity criterion)
AB/PQ = BD/QM = AD/PM