Question

Sides $\mathrm{AB}$ and $\mathrm{BC}$ and median $\mathrm{AD}$ of a triangle $\mathrm{ABC}$ are respectively proportional to sides $\mathrm{PQ}$ and $\mathrm{QR}$ and median $\mathrm{PM}$ of $\mathrm{\Delta PQR}$. Show that $\mathrm{\Delta ABC}~\mathrm{\Delta PQR}$.

Answer 1

Determine to prove that $\mathrm{\Delta ABC}~\mathrm{\Delta PQR}$.

Given that $\mathrm{\Delta ABC}$ and $\mathrm{\Delta PQR},\mathrm{AB},\mathrm{BC}$ and median $\mathrm{AD}$ of $\mathrm{\Delta ABC}$ are proportional to sides $\mathrm{PQ},\mathrm{QR}$ and median $\mathrm{PM}$ of $\mathrm{\Delta PQR}$.

From given conditions

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

We know that

$$\begin{gathered} \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}} \\ \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{{\displaystyle \frac{1}{2}}\mathrm{BC}}{{\displaystyle \frac{1}{2}}\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}} \end{gathered}$$

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{BD}{QM}=\frac{\mathrm{AD}}{\mathrm{PM}}$ ($\mathrm{D}$ is the midpoint of $\mathrm{BC}$. $\mathrm{M}$ is the midpoint of $\mathrm{QR}$)

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

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

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

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

$$\begin{gathered} \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\dots \dots \dots \dots \dots \dots \dots \left(\mathrm{i}\right) \\ \angle \mathrm{ABC}=\angle \mathrm{PQR}..\dots \dots \dots \dots \dots \dots \left(\mathrm{ii}\right) \end{gathered}$$

From equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get,

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

Hence, it is proved as $\mathbf{\Delta ABC}\mathbf{}\mathbf{~}\mathbf{}\mathbf{\Delta PQR}$.