Question

If AD and PM are medians of ΔABC and ΔPQR respectively, where ΔABC ~ ΔPQR; prove that $\frac{AB}{PQ}=\frac{AD}{PM}$.

Answer 1

Since, AD and PM are the medians of ΔABC and ΔPQR respectively
Therefore, BD = DC = $\frac{BC}{2}$ and QM = MR = $\frac{QR}{2}$ ...(1)

Now,
ΔABC ~ ΔPQR

As we know, corresponding sides of similar triangles are proportional.
Thus, $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$ ...(2)

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

From (1) and (2), we get
$$\begin{gathered} \frac{AB}{PQ}=\frac{BC}{QR} \\ \Rightarrow \frac{AB}{PQ}=\frac{2BD}{2QM} \\ \Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}...\left(4\right) \end{gathered}$$

Now, in ΔABD and ΔPQM

$$\begin{gathered} \frac{AB}{PQ}=\frac{BD}{QM}\left(\mathrm{From}\left(4\right)\right) \\ \angle B=\angle Q\left(\mathrm{From}\left(3\right)\right) \end{gathered}$$

By SAS similarity,
ΔABD ~ ΔPQM

Therefore, $\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$.

Hence, $\frac{AB}{PQ}=\frac{AD}{PM}$.