# Q14  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 $\Delta ABC \sim \Delta PQR$

S seema garhwal

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$              (given)

Produce AD and PM to E and L such that AD=DE  and PM=DE. Now,

join B to E,C to E,Q to L and R to L.

AD and PM are medians of a triangle, therefore

QM=MR  and    BD=DC

PM=ML               (By construction)

So, diagonals of ABEC bisecting each other at D,so ABEC is a parallelogram.

Similarly, PQLR is also a parallelogram.

Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$               (Given )

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{2.AD}{2.PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}$

$\Delta ABE \sim \Delta PQL$                (SSS similarity)

$\angle BAE=\angle QPL$ ...................1           (Corresponding angles of similar triangles)

Similarity, $\triangle AEC=\triangle PLR$

$\angle CAE=\angle RPL$........................2

$\angle BAE+\angle CAE=\angle QPL+\angle RPL$

$\angle CAB=\angle RPQ$............................3

In $\triangle ABC\, and\, \, \triangle PQR,$

$\frac{AB}{PQ}=\frac{AC}{PR}$        ( Given )

$\angle CAB=\angle RPQ$        ( From above equation 3)

$\triangle ABC\sim \triangle PQR$         ( SAS similarity)

Exams
Articles
Questions