Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \(\triangle\) ABC ~ \(\triangle\) PQR.
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 …