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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

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

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

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