well-known geometry

perfect_radio

Let $E,F$ be the feet of the perpendiculars from the vertices $B,C$ of triangle $\triangle ABC$ . Let $O$ be the circumcenter $\triangle ABC$ . Prove that $OA \perp FE .$

**Posted:** Mon Jul 25, 2005 12:52 pm

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socrates · Yang-Mills Theory Offline Joined: 29 Jun 2005 Posts: 740

Extend $AO$ to meet the circumcircle in $Z$ .
See also that the quadrilateral $EFCB$ is inscriptible and finally show that
$ZAC+AFE=90^0$

It is a good theorem

If I am not mistaken, this is due to $Nagel$ ....... Smile

**Posted:** Tue Jul 26, 2005 3:52 am

riddler

**Posted:** Tue Jul 26, 2005 4:20 am

Arne

He was a mathematician/geometer who found some interesting theorems...

The Nagel point of a triangle is attributed to him. (This point is the intersection of the three cevians connecting a vertex with the tangency point of the excircle touching the opposite side.)

**Posted:** Tue Jul 26, 2005 4:38 am

Hanno

Hi all.

Let H be the intersection of $EB$ and $DF$ and let $AG$ be perpendicular to $BC$ through $A$ . Then, since $\angle HDA=\angle AFH=90°$ , the quadrilateral $AEHF$ is 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.
One may also notice that the triangles $AEF$ and $ABC$ are similiar.

Hanno

**Posted:** Tue Jul 26, 2005 5:30 am