# Quadrilaterals and parallel lines: An olympiad question

Let $$ABCD$$ be a convex quadrilateral $$\measuredangle ADC = \measuredangle BCD > 90$$. Let $$E$$ be the point in which line $$AC$$ intersects the line parallel to $$AD$$ through $$B$$ and Let $$F$$ be the point in which line $$BD$$ intersects line parallel to $$BC$$ through $$A$$. Prove $$EF||CD$$.

I have tried multiple ways to prove this but am not arriving at the proof. Kindly give some hint or help me in solving this question

Denote the intersection of $$\overline{AC}$$ and $$\overline{BD}$$ by $$O$$. Since $$\overline{AD}||\overline{BE}$$,

$$\overline{DO}:\overline{OB} = \overline{AO}:\overline{OE}.$$

Since $$\overline{BC}||\overline{AF}$$,

$$\overline{BO}:\overline{OF} = \overline{CO}:\overline{OA}.$$

Multiplying the ratios, $$\overline{DO}:\overline{OF} = \overline{CO}:\overline{OE}.$$

• Thanks a lot. Got it – ShiS Sep 24 '18 at 17:55
• How do you know trhat $BC$ is parallel to $AF$? – Mike Sep 24 '18 at 18:27
• That is the assumption. – Hw Chu Sep 24 '18 at 18:28
• My mistake you are correct – Mike Sep 24 '18 at 18:33

Nonstandard, simple, a bit over powered, but most creative solution:

Consider a homothety $$H_1$$ with center at $$O$$ which takes $$A\mapsto C$$. Then it takes $$F\mapsto B$$.

Also consider a homothety $$H_2$$ with center at $$O$$ which takes $$E\mapsto A$$. Then it takes $$B\mapsto D$$.

Then composition $$H_2\circ H_1$$ takes $$F\to B$$ and composition $$H_1\circ H_2$$ takes $$E\to C$$.

Now since $$H_1$$ and $$H_2$$ have the same center they comute: $$H_1\circ H_2= H_2\circ H_1$$ and name this composition with $$H$$. So $$H$$ takes $$F\mapsto B$$ and $$E\mapsto C$$ so it takes line $$EF$$ to line $$BC$$, so $$EF||BC$$.