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Let $ABC$ be a right angled triangle, where the right angle is at $A$. Construct squares on $AC$, $AB$ and $BC$ as shown. Let $P$ be the point of intersection of $BK$ and $FC$ (Note that $P$ is not marked in the figure).

Then I conjecture that $AP$ is parallel to $BD$.

enter image description here

What I tried:

By observing that $\Delta FBC\cong \Delta ABD$, we see that $\angle BAC=\angle BFC$. Therefore, if $X$ is the point of intersection of $FC$ and $AD$, we see that $BFAX$ is a cyclic quadrilateral.

This gives us that $AD\perp FC$, and similarly $BK\perp AE$. But I couldn't go any further.

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Fig.1

By considering the congruent (and similar) right angle triangles it is easy to prove (similar to as you did) that $BK\bot MC$, $CF\bot BM$ and $MA\bot BC$. All three height of the triangle $MBC$ cross in one point ($=P$), thus $AP\bot BC$.

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  • $\begingroup$ This seems great. I still haven't figured out why $MP$ should pass through $A$ though. $\endgroup$ – caffeinemachine Aug 11 '15 at 13:03
  • $\begingroup$ There is no $P$ from the beginning. We take three heights $BK, CF$ and $MA$ (continued to the side $BC$). They must cross in one point. Call the point $P$. So $MA$ goes though the cross-point of two green lines and is orthogonal to $BC$. $\endgroup$ – A.Γ. Aug 11 '15 at 13:06
  • $\begingroup$ Forget my comment. It was a bit hasty. Thanks. $\endgroup$ – caffeinemachine Aug 11 '15 at 13:09
  • $\begingroup$ @caffeinemachine You are welcome, and thanks for the nice problem. $\endgroup$ – A.Γ. Aug 11 '15 at 13:11
  • $\begingroup$ No. Thanks for the masterful solution! $\endgroup$ – caffeinemachine Aug 11 '15 at 13:46

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