Quadrilaterals and parallel lines: An olympiad question

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

Answer 1

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

Answer 2

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

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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 $B\mapsto D$.

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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 $EF||BC$.