Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
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