Problems in triangles

Question

Proof : In an acute angle triangle $ABC$, $AP$ is the altitude. Circle drawn with $AP$ as its diameter cut the side $AB$ and $AC$ at $D$ and $E$, respectively then length of $DE$ is equal to (Area of Triangle) /Circumradius

Answer 1

$$\measuredangle AED=\measuredangle APD=90^{\circ}-\measuredangle BPD=\measuredangle B,$$ which says that $$\Delta ABC\sim\Delta AED.$$

Thus, $$\frac{DE}{BC}=\frac{AD}{AC},$$ which says $$DE=\frac{BC\cdot AD}{AC}=\frac{BC\cdot AP\sin\measuredangle B}{2R\sin\measuredangle B}=\frac{S_{\Delta ABC}}{R}.$$