Question

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectively where $\Delta ABC~\Delta PQR$ prove that $\frac{AB}{PQ}=\frac{AD}{PM}$.

Answer 1

Step1: Given and prove that

1. $\Delta ABC~\Delta PQR$
2. $AD$ and $PM$ are medians of triangles $ABC$and $PQR$ respectively.

Prove that: $\frac{AB}{PQ}=\frac{AD}{PM}$

Step 2: Proof

The corresponding sides of similar triangles are in proportion.

$\Rightarrow$ $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\dots \dots \dots \dots \dots \dots \dots \dots \left(i\right)$

$\Rightarrow \angle A=\angle P,\angle B=\angle Q,\angle C=\angle R\dots \dots \dots \dots .\dots ..\left(ii\right)$

Since, $AD$ and $PM$ are medians, they will divide the opposite sides of the triangles.

$\therefore BD=\frac{BC}{2}$

$\therefore QM=\frac{QR}{2}\dots \dots ..\dots \dots \dots \dots .\left(iii\right)$

From equations$\left(i\right)$ and $\left(iii\right)$, we get

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}\dots \dots \dots \dots \dots \dots \dots .\left(iv\right)$

In $\Delta ABD$ and $\Delta PQM$,

$\Rightarrow \angle B=\angle Q$ [from equation $\left(ii\right)$]

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}$ [from equation $\left(iv\right)$]

$\therefore \Delta ABD~\Delta PQM$ [SAS similarity criterion]

Since both the triangles $\Delta ABD$ and $\Delta PQM$ are similar, the corresponding sides of the similar triangle are in proportion.

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

Hence, proved.