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

Solution :

Given : AD and PM are the medians of triangles ABC and PQR respectively, where triangles\(\triangle\) ABC ~ \(\triangle\) PQR.

To Prove : \(AB\over PQ\) = \(AD\over PM\)

Proof  : In triangles ABD and PQM, we have

\(\angle\) B = \(\angle\) D        (because \(\triangle\) ABC ~ \(\triangle\) PQR)

Since AD and PM are the medians to BC and QR respectively and \(AB\over PQ\) = \(BC\over QR\)

\(AB\over PQ\) = \({1\over2}BC\over {1\over 2}QR\)

So, \(AB\over PQ\) = \(BD\over QM\)

\(\therefore\)  By SAS similarity,

\(\triangle\) ABD ~ \(\triangle\) PQM

So, \(AB\over PQ\) = \(BD\over QM\) = \(AD\over PM\)

Leave a Comment

Your email address will not be published.