Question

In the figure, given below, $$2AD=AB, P$$ is mid-point of $$AB, Q$$ is mid-point of $$DR$$ and $$PR\parallel BS$$. Prove that:
$$DS\parallel 3RS$$

Answer 1

From triangle $$ABC, P$$ is the midpoint and $$PR\parallel BS$$
Therefore $$R$$ is the midpoint of $$BC$$
From $$\triangle BRS$$ and $$\triangle QRC$$
$$\angle BRS=\angle QRC$$
$$BR=RC$$
$$\angle RBS=\angle RCQ$$
$$\therefore \triangle BRS\cong \triangle QRC$$
$$\therefore QR=RS$$
$$DS=DQ+QR+RS=QR+QR+RS=3RS$$