If $A D$ and $P M$ are medians of $\triangle A B C$ and $\triangle P Q R$ respectively, where $\triangle A B C \sim \triangle P Q R ;$ prove that $\frac{A B}{P Q}=\frac{A D}{P M}$.
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, $\frac{A B}{P Q}=\frac{A D}{P M}$.