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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

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  • $\begingroup$ We need something more in the given. $\endgroup$ – Michael Rozenberg Jan 1 '18 at 13:09
  • $\begingroup$ I have added the image $\endgroup$ – Pokemon Ash Jan 1 '18 at 13:17
  • $\begingroup$ This is January 1, not April 1. $\endgroup$ – Professor Vector Jan 1 '18 at 13:20
  • $\begingroup$ @Pokemon Ash I think your picture drown for another problem. $\endgroup$ – Michael Rozenberg Jan 1 '18 at 13:23
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$$\measuredangle AED=\measuredangle APD=90^{\circ}-\measuredangle BPD=\measuredangle B,$$ which says that $$\Delta ABC\sim\Delta AED.$$

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

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  • $\begingroup$ You're referring to the image the OP meant to give, not the one they gave, right? ;) $\endgroup$ – Professor Vector Jan 1 '18 at 13:23
  • $\begingroup$ Yes, of course. Happy New Year! :) $\endgroup$ – Michael Rozenberg Jan 1 '18 at 13:25
  • $\begingroup$ Sorry for the mistake $\endgroup$ – Pokemon Ash Jan 1 '18 at 13:35
  • $\begingroup$ It's all right. Happy New Year! $\endgroup$ – Michael Rozenberg Jan 1 '18 at 14:02

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