Question

Sides $\mathrm{AB},\mathrm{BC}$ and Median $\mathrm{AD}$ of a $\mathrm{\Delta ABC}$ are proportional to sides $\mathrm{PQ},\mathrm{QR}$ and median $\mathrm{PM}$of $\mathrm{\Delta PQR}$ . Prove $∆\mathrm{ABC}~∆\mathrm{PQR}$ .

Answer 1

Step 1.$\mathrm{\Delta ABD}~\mathrm{\Delta PQM}$ :

Two triangles. $\mathrm{\Delta }\mathrm{ABC}$ and $\mathrm{\Delta PQR}$ in which $\mathrm{AB},\mathrm{BC}$ and median $\mathrm{AD}$ of $\mathrm{\Delta ABC}$ is proportional to sides$\mathrm{PQ},\mathrm{QR}$ and median $\mathrm{PM}\mathrm{of}\mathrm{\Delta PQR}$

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

Since,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$ ($D$ is the mid-point of $BC$ , $M$ is the midpoint of $QR$)

$\mathrm{\Delta ABD}~\mathrm{\Delta PQM}$ [$SSS$ similarity criterion]

Step 2. $∆\mathrm{ABC}~∆\mathrm{PQR}$ :

Therefore, $\angle \mathrm{ABD}=\angle \mathrm{PQM}$ [Corresponding angles of two similar triangles are equal]

$\angle \mathrm{ABC}=\angle \mathrm{PQR}$

In $\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta PQR}$

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}...\left(i\right)$

$\angle \mathrm{ABC}=\angle \mathrm{PQR}...\left(ii\right)$)

Using the above equations $\left(i\right)$ and , we get

$\mathrm{\Delta ABC}~\mathrm{\Delta PQR}$ [By SAS similarity criterion]

Hence, it is proven that $∆\mathrm{ABC}~∆\mathrm{PQR}$