Question

The side $$AC$$ of a triangle $$ABC$$ is produced to point $$E$$ so that $$CE=\dfrac{1}{2}AC.\ D$$ is the mid-point of $$BC$$ and $$ED$$ produced meets $$AB$$ at $$F$$. Lines through $$D$$ and $$C$$ are drawn parallel to $$AB$$ which meet $$AC$$ at point $$R$$ respectively. Prove that:
$$4CR=AB$$

Answer 1

Given that $$D$$ is the midpoint of $$BC$$ and $$DP$$ is parallel to $$AB$$, therefore $$P$$ is the midpoint of $$AC$$
$$PD=\dfrac{1}{2}AB$$
From the triangle $$PED$$ we have $$PD\parallel CR$$ and $$C$$ is the midpoint of $$PE$$ therefore $$CR=\dfrac{1}{2}PD$$
$$PD=\dfrac{1}{2}AB$$
$$\dfrac{1}{2}PD=\dfrac{1}{4}AB$$
$$CR=\dfrac{1}{4}AB$$
$$4CR=AB$$
Hence it is shown.