I think I've solved the following problem, but I don't know whether my proof is valid. I would appreciate it if someone could verify it for me. Moreover, I feel as if the solution shouldn't be as convoluted, so a shorter/simpler one is also welcomed. Thanks in advance!
In the following figure, assume BY=CZ are altitudes. Show that AC=AB.
Proof. Consider the right triangles $\triangle BCY$ and $\triangle CBZ$. Since they share BC as their hypothenuse, and the arm $BY$ is equal to the arm $CZ$ (by assumption), then, by the Hypothenuse-Leg postulate, they are congruent.
Notice that $\angle YBC$ and $\angle ZCB$ are corresponding, and let $O$ be the point where $BY$ and $CZ$ meet. Then, by the Pons Asinorum, $\triangle BOC$ is isosceles, and $BO=CO$. Then $YO=BY-BO=CZ-CO=OZ$, so $YO=ZO$.
Now draw line $AO$, and consider right triangles $\triangle OZA$ and $\triangle OYA$. Since they share AO as their hypothenuse, and arm $YO$ is equal to arm $ZO$, then, by the Hypothenuse-Leg postulate, they are congruent. Then, $AZ=AY\implies AZ+BZ=AY+CY\implies AB=AC,$ as desired.
i would write $$A=\frac{1}{2}bh_b=\frac{1}{2}ch_c$$ since $$h_c=h_b$$ we get $$b=c$$