TITLE: A Fact I Observed While Looking at the Proof of Pythagorean Theorem.
QUESTION [9 upvotes]: 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 $BD$.


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.

REPLY [6 votes]: 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 $AP\bot BC$.