Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of \(\triangle\) PQR (see Fig 6.41). Show that \(\triangle\) ABC ~ \(\triangle\) PQR.

Given, \(\triangle\) ABC and \(\triangle\) PQR, AB, BC and median AD of \(\triangle\) ABC are proportional to sides PQ, QR and median PM of \(\triangle\) PQR

i.e. AB/PQ = BC/QR = AD/PM

We have to prove: \(\triangle\) ABC ~ \(\triangle\) PQR

As we know here,
AB/PQ = BC/QR = AD/PM
\({{AB} \over {PQ}} = {{{{1} \over {2}}BC} \over {{{1} \over {2}}QR}} = {{AD} \over {PM}}\)
AB/PQ = BC/QR = AD/PM
(D is the midpoint of BC. M is the midpoint of QR)
\(\triangle\) ABD ~ \(\triangle\) PQM [SSS similarity criterion]

\(\angle\) ABD = \(\angle\) PQM
[Corresponding angles of two similar triangles are equal]
\(\angle\) ABC = \(\angle\) PQR

In \(\triangle\) ABC and \(\triangle\) PQR
AB/PQ = BC/QR ………………………….(i)
\(\angle\) ABC = \(\angle\) PQR ……………………………(ii)

From equation (i) and (ii), we get,
\(\triangle\) ABC ~ \(\triangle\) PQR [SAS similarity criterion]