The points $D$ and $E$ on the circumcircle of an acute triangle $ABC$ are such that $AD=AE = BC$. Let $H$ be the common point of the altitudes of triangle $ABC$.
It is known that $AH^{2}=BH^{2}+CH^{2}$.
Prove that $H$ lies on the segment $DE$.
Proposed by D. Shiryaev
The missing condition is AD = AE = BC. Power of the orthocenter H to (O, R) is $8R^{2}\cos A \cos B \cos C = 4R^{2}(1-(\cos^{2}A+\cos^{2}B+\cos^{2}C)),$ while power of H to (A, a) is $a^{2}-4R^{2}\cos^{2}A = 4R^{2}(1-2 \cos^{2}A).$ These are equal (H lies on the radical axis DE of (O), (A)) iff
$1-2 \cos^{2}A = 1-(\cos^{2}A+\cos^{2}B+\cos^{2}C)\ \Longleftrightarrow$
$\cos^{2}A = \cos^{2}B+\cos^{2}C\ \Longleftrightarrow\ AH^{2}= BH^{2}+CH^{2}$