Y by Adventure10, Mango247
Let
be the feet of the perpendiculars from the vertices
of triangle
. Let
be the circumcenter
. Prove that ![\[ OA \perp FE .\]](//latex.artofproblemsolving.com/a/9/d/a9db29e528d86ad6b58f68ccfca298245f167894.png)





![\[ OA \perp FE .\]](http://latex.artofproblemsolving.com/a/9/d/a9db29e528d86ad6b58f68ccfca298245f167894.png)
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EBand
DFand let
AGbe perpendicular to
BCthrough
A. Then, since
\angle HDA=\angle AFH=90°, the quadrilateral
AEHFis cyclic. Thus, we have
\angle AFE=\angle AHE=\angle GHB=90°-\angle EBC=\angle BCA. Finally, since
\angle HAE=90°-\frac{1}{2}\angle BHA=90°-\angle BCA, the conclusion follows.
AEFand
ABCare similiar.
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