If AD and PM are medians of ΔABC and ΔPQR respectively, where ΔABC ~ ΔPQR; prove that

Solution:

Since, AD and PM are the medians of ΔABC and ΔPQR respectively

Therefore, $B D=D C=\frac{B C}{2}$ and $Q M=M R=\frac{Q R}{2}$      ...(1)

Now,
ΔABC ~ ΔPQR

As we know, corresponding sides of similar triangles are proportional.

Thus, $\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}$        ...(2)

Also, $\angle A=\angle P, \angle B=\angle Q$ and $\angle C=\angle R$      ...(3)

From (1) and (2), we get

$\frac{A B}{P Q}=\frac{B C}{Q R}$

$\Rightarrow \frac{A B}{P Q}=\frac{2 B D}{2 Q M}$

$\Rightarrow \frac{A B}{P Q}=\frac{B D}{Q M} \quad \ldots(4)$

Now, in ΔABD and ΔPQM

$\frac{A B}{P Q}=\frac{B D}{Q M}$       (From (4))

$\angle B=\angle Q$     (From (3))

By SAS similarity,
ΔABD ~ ΔPQM

Therefore, $\frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}$.

Hence, $\frac{A B}{P Q}=\frac{A D}{P M}$.