# A Fact I Observed While Looking at the Proof of Pythagorean Theorem.

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.

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 $AP\bot BC$.
• This seems great. I still haven't figured out why $MP$ should pass through $A$ though. – caffeinemachine Aug 11 '15 at 13:03
• 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 $BC$. – A.Γ. Aug 11 '15 at 13:06