Proof. $ \left\{\begin{array}{c}PA^{2}=AC\cdot AE=bc\cdot\cos A\\\ QA^{2}=AB\cdot AF=cb\cdot \cos A\end{array}\right\|\implies PA=QA=\sqrt{\frac{b^{2}+c^{2}-a^{2}}{2}}$.

Virgil Nicula wrote: Proof. $ \left\{\begin{array}{c}PA^{2} = AC\cdot AE = bc\cdot\cos A \\ \ QA^{2} = AB\cdot AF = cb\cdot \cos A\end{array}\right\|\implies PA = QA = \sqrt {\frac {b^{2} + c^{2} - a^{2}}{2}}$. Why is it exactly that $ PA^2 = AC \cdot AE$? I understand the rest but just not that part.

modularmarc101 wrote: Why is it exactly that $ PA^2 = AC \cdot AE$? I understand the rest but just not that part. [geogebra]2806dcaf8f8f8944f26801071af9694516c9a14f[/geogebra] Since $ PE$ is the altitude in right $ \triangle APC$, so by similarity $ \frac {PA}{PC} = \frac {AE}{PA}$ so $ PA^2 = AC \cdot AE$.

It's easy to see and worth mentioning that, $E,F$ are nothing but the foot of perpendiculars from $B,C$, Now, Let $D$ be the foot of altitude from $A$, and Construct a circle $\omega$ with radius $\sqrt{AH \cdot AD}$ Lemma: Let $\omega \cap \odot (ABC)=X,Y$, then, $X,Y$ lie on $EF$ Proof: Perform $\Phi$ (inversion) around $\omega$, and note that $A,X,Y$ remain invariant under $\Phi$, of course, $\Phi : B \longleftrightarrow F$ $ ; C \longleftrightarrow E$ which implies $XYCB$ is cyclic $\implies $ $X,Y \in EF$ $\qquad \blacksquare$ Easy to see, that both, $C_1,C_2$ pass through $D$ in the original diagram, $\Phi :$ $D$ $\mapsto$ $H$, $\longrightarrow \text{Orthocenter}$ in $\Delta ABC$, Note that $\Phi : C_1 \mapsto CF , C_2 \mapsto BE$, Denote $P^*,Q^*$ as the image of $P,Q$ under $\Phi$, then, notice, $Q$ already lies on $C_1 $ and $CF$, and $Q^*$ will also lie on the intersection of $C_1$ and $CF$ which is $Q$, hence, $P \equiv P^*$ and $Q \equiv Q^*$, hence, $P,Q$ lie on $\omega$ $\implies$ $AP=AQ$