Question

If two sides and median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle then prove that the triangles are similar.

Answer 1

Let us extend $AD$ to point $D$ such that $AD=DE$ and $PM$ up to point $L$ such that $PM=ML$

Join $B$ to $E,C$ to $E,Q$ to $L$ and $R$ to $L$

We know that medians are the bisector of the opposite side

Hence $BD=DC$

Also,$AD=DE$ (by construction)Hence in quadrilateral $ABEC$, diagonals $AE$ and $BC$ bisect each other at point $D$

$\therefore$ quadrilateral $ABEC$ is a parallelogram.

$\therefore AC=BE$ and $AB=EC$ .......$\left(1\right)$ since opposite sides of a parallelogram are equal.

We know that medians are the bisector of the opposite side

Hence $QM=MR$

Also, $PM=ML$ (by construction)Hence in quadrilateral $PQLR$, diagonals $PL$ and $QR$ bisect each other at point $M$

$\therefore$ quadrilateral $PQLR$ is a parallelogram.

$\therefore PR=RL$ and $PQ=QL$ ........$\left(1\right)$since opposite sides of a parallelogram are equal.

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

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AD}{PM}$ from $\left(1\right)$ and $\left(2\right)$

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

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

as $AD=DE$

$AE=AD+DE=AD+AD=2AD$

$PM=ML$

$PL=PM+ML=PM+PM=2PM$

$\therefore △ABE\sim △PQL$ by SSS similarity criterion

$\therefore △ABE\sim △PQL$ and $△AEC\sim △PLR$

We know that corresponding angles of similar triangles are equal

$\therefore \angle BAE=\angle QPL$ ........$\left(3\right)$

and $\angle CAE=\angle RPL$ ........$\left(4\right)$

Adding $\left(3\right)$ and $\left(4\right)$ we get

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

$\angle CAB=\angle RPQ$ ........$\left(5\right)$

In $△ABC$ and $△PQR$

$\frac{AB}{PQ}=\frac{AC}{PR}$

$\angle CAB=\angle RPQ$

$\therefore △ABC\sim △PQR$ by SAS similarity criterion

Hence proved.