This is actually a particular case of the Butterfly Theorem.

http://www.cut-the-knot.org/pythagoras/Butterfly.shtml K_Er_Tus's solution is correct !Thank you !

Let $ K,\ L$ be, the orthogonal projections of $ B,\ C$ respectively, on the line segment $ EF$ and we denote the points $ X\equiv BC\cap AH,\ Y\equiv AC\cap BH,\ Z\equiv AB\cap CH.$ It is easy to show that $ HK = HL,$ from $ MB = MC$ and $ BK\parallel MH\parallel CL$ and then, the points $ B'\equiv CL\cap BH$ and $ C'\equiv BK\cap CH,$ are the reflexions of $ B,\ C$ respectively, with respect to $ H.$ We have now, the configuration of two congruent triangles $ \bigtriangleup HBC',\ \bigtriangleup HB'C,$ with $ E,\ F$ as their orthocenters respectively and so, their distances from their homologous vertices, are equal. Hence, we conclude that $ EH = FH$ and the proof is completed. Kostas Vittas.

Great vittasko !

Thank you very much. I like this problem. It's easy but we can solve it by many ways. Use Butter fly Theorem is a nice example.

K_Er_Tus wrote: This is actually a particular case of the Butterfly Theorem. thanhnam2902 wrote: Thank you very much. I like this problem. It's easy but we can solve it by many ways. Use Butter fly Theorem is a nice example. Can someone explain to me what Butterfly theorem is?

oh! It is very famous Theorem. I have posted it at this forum.

You see it at here: http://www.cut-the-knot.org/pythagoras/Butterfly.shtml

Thank you so much!. I realized that it is a particular case of a more general result: $AB$ is a chord of a circle $(O)$ and $P$ is a point on $AB.$ Two distintc chords $MN$ and $RS$ pass through $P.$ $MS$ and $RN$ cut $AB$ at $X,Y,$ respectively. Then we have \[ \frac{1}{PA}-\frac{1}{PX}=\frac{1}{PB}-\frac{1}{PY} \]

Let $ M'$ the diametrically oposite point of $ A$ in the circumcircle of $ \triangle{ABC}$. $ H,M,M'$ are collinear since by angle chasing $ BHCM'$ is a parallelogram. Also, $ EHM'B$ and $ HFCM'$ are both cyclic (opposite angles equal $ \frac{\pi}{2}$, sum $ \pi$. Since $ \angle{HBE}=\angle{HCF}$ this implies $ HM$ is angle bisector of $ EM'F$ then $ EH=HF$.

Nemion wrote: Let $ M'$ the diametrically oposite point of $ A$ in the circumcircle of $ \triangle{ABC}$. $ H,M,M'$ are collinear since by angle chasing $ BHCM'$ is a parallelogram. Also, $ EHM'B$ and $ HFCM'$ are both cyclic (opposite angles equal $ \frac {\pi}{2}$, sum $ \pi$. Since $ \angle{HBE} = \angle{HCF}$ this implies $ HM$ is angle bisector of $ EM'F$ then $ EH = HF$. This is exactly what I am thinking about

Dear Mathlinkers, for the butterfly theorem you can see a nice article written by Darij Grinberg http://www.cip.ifi.lmu.de/~grinberg/ On cyclic quadrilaterals and the butterfly theorem Sincerely Jean-Louis

Two orthogonal lines through orthocenter is the signature of Droz-Farny configuration. The required equality of segments follows immediately. M.T.

If N=BH∩AC and P=CH∩AB, then BCNP is cyclic, M is it’s excenter and, from butterfly theorem, EH = FH. Best regards, sunken rock