# Problems in triangles

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

• We need something more in the given. – Michael Rozenberg Jan 1 '18 at 13:09
• I have added the image – Pokemon Ash Jan 1 '18 at 13:17
• This is January 1, not April 1. – Professor Vector Jan 1 '18 at 13:20
• @Pokemon Ash I think your picture drown for another problem. – Michael Rozenberg Jan 1 '18 at 13:23

$$\measuredangle AED=\measuredangle APD=90^{\circ}-\measuredangle BPD=\measuredangle B,$$ which says that $$\Delta ABC\sim\Delta AED.$$
Thus, $$\frac{DE}{BC}=\frac{AD}{AC},$$ which says $$DE=\frac{BC\cdot AD}{AC}=\frac{BC\cdot AP\sin\measuredangle B}{2R\sin\measuredangle B}=\frac{S_{\Delta ABC}}{R}.$$