Prove that $H$ is the orthocenter of $\Delta ABC$. 
In acute $\Delta ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively such that $B, C, D, E$ are concyclic, $BE$ and $CD$ intersect at $H$, and $H$ is on the altitude of $\Delta ABC$ passing through $A$. Prove that $H$ is the orthocenter of $\Delta ABC$.

I tried using similar triangles, Pythagorean theorem, Ceva's theorem, but all failed. Please help. Thank you.
 A: The (final) statement is not true.
Consider an acute isosceles triangle with vertex angle $A$. Because the figure is symmetric about the altitude through $A$, any circle through $B$ and $C$ that meets $AB$ and $AC$ will meet these edges at points $D$ and $E$ such that $BD$ and $CE$ have intersection $H$ somewhere on that altitude.
On one extreme, the smallest circle under consideration is the one with diameter $BC$, which in fact leads to orthocenter $H$ (since $\angle BDC$ and $\angle BEC$ subtend a semicircle); on the other, the largest circle is the circumcircle of $\triangle ABC$, for which we have $H=A$. Other circles lead to points in between the orthocenter and $A$, so the conditions do not "usually" cause $H$ to be the orthocenter.

A note about the acuteness condition: were $\triangle ABC$ a right isosceles triangle, the two "extreme" circles mentioned above would coincide, and we would indeed have $H=A$ lie at the orthocenter. Were the triangle obtuse, then it would lie entirely within the smallest circle (with diameter $BC$), allowing no intersection points $D$ and $E$ with the sides of the triangle. (Extended sides are another matter.)
A: 
Theorem: Let $O$ be a circle with chord $AB$. Let $\widehat{AB}$ denote an arc of $O$ ending on $A$ and $B$. Choose $C$ and $D$ on $\widehat{AB}$ such that $C$ is adjacent to $B$ (and $D$ is adjacent to $A$) and let $AD$ intersect $BC$ at $K$. Let ${AC}$ intersect $BD$ at $H$. Then $H$ lies on the perpendicular of $AB$ through $K$ if and only if $AB$ is a diameter of $O$ or $AK = BK$.
Proof: Suppose the perpendicular of $AB$ through $K$ intersects $AB$ at $G$. Then $KG$, $AC$ and $BD$ are cevians of the triangle. By the trigonometric form of Ceva's theorem, the three are concurrent if and only if 
$$\sin(\angle KBD)\sin(\angle BAC)\sin(\angle AKG) = \sin(\angle ABD)\sin(\angle KAC)\sin(\angle BKG)$$
Let us now compile some angular equivalencies present in $\triangle ABK$. First, note that $$\angle KAC = \angle KBD$$
since they both subtend $\widehat{CD}$. By a bit of angle chasing, we also have
$$\angle ABD = (90 - \angle ADB) + \angle AKG$$
$$\angle BAC = (90 - \angle ACB) + \angle BKG$$
Again, notice that $\angle ADB = \angle ACB$ since they both subtend $\widehat{AB}$. Let $x$ denote $90 - \angle ACB$. From this, Ceva's theorem simplifies as
$$\sin(x + \angle BKG)\sin(\angle AKG) = \sin(x + \angle AKG)\sin(\angle BKG)$$
expanding the sines using angle addition identities, the equation reduces to
$$\sin(x)\cos(\angle BKG)\sin(\angle AKG) = \sin(x)\cos(\angle AKG)\sin(\angle BKG)$$
or equivalently
$$\sin(x)\tan(\angle AKG) = \sin(x)\tan(\angle BKG)$$
clearly we have equality if and only if $\sin(x) = 0$ which implies 
$$\angle ACB = \angle ADB = 90$$ 
This of course implies that $AB$ is a diameter of $O$. Or we have 
$$\tan(\angle AKG) = \tan(\angle BKG) \implies \angle AKG = \angle BKG$$
in which case $\triangle AKB$ is isosceles with $AK = BK$. $\square$
Now, suppose that the hypothesis of your proposition is satisfied. Let $O$ denote the circumcircle of $BCDE$ of which $BC$ is a chord. Since $H$ lies on the altitude through $A$, by the above theorem, either triangle $\triangle ABC$ is isosceles or $BC$ is a diameter of $O$. If the latter, then $CD \perp AB$ and $BE \perp AC$ so that $H$ is indeed the orthocenter of the triangle. Note that the condition that $\triangle ABC$ is acute is not needed.
