Solution :
Given : \(\triangle\) ABC and \(\triangle\) PQR in which AD and PM are the medians, such that
\(AB\over PQ\) = \(BC\over QR\) = \(AD\over PM\)
To prove : \(\triangle\) ABC ~ \(\triangle\) PQR
Proof : \(AB\over PQ\) = \(BC\over QR\) = \(AD\over PM\)
\(\implies\) \(AB\over PQ\) = \({1\over 2}BC\over {1\over 2}QR\) = \(AD\over PM\)
\(\implies\) \(AB\over PQ\) = \(BD\over QM\) = \(AD\over PM\)
By SSS similarity,
\(\triangle\) ABD ~ \(\triangle\) PQM
\(\implies\) \(\angle\) B = \(\angle\) Q
Now, in triangle ABC and PQR, we have
\(AB\over PQ\) = \(BC\over QR\) (given)
and \(\angle\) B = \(\angle\) Q
So, By SAS similarity,
\(\triangle\) ABC ~ \(\triangle\) PQR