This problem is not hard, actually I solved it in 10 minutes

yup...simply putting coordinates making $B$ and $C$ symmetric about the y-axis gives the proof in 2 lines.

Note that $H$ is orthocenter in $BGD$, if $G$ is the intersection of the medians. Other possible solution is to trig/length bash, it works very nicely using the perpendicularity condition (consequence of Pythagorean theorem).

My solution: Let $G$ be the centroid of $ABC$. Let $T \in BC$ such that $CD=DT$. $I$ is the midpoint of $BC$. $DH $ cuts $BE$ at $F$ Then it's easy to prove that $IH.IA = IT.IB$, so $H,A,T,B$ are cyclic, then we have the desired result

Let $AH\cap BE=K, AH\cap BC=F$ and let $H'$ be the reflection of $H$ wrt $BC$. $3|CD|=|BC|\Rightarrow \frac{|BD|}{|DC|}=\frac 21$. We have $|AB|=|AC|$ and $AH\perp BC$. Hence, $AH$ is the angle bisector and median. By the angle bisector theorem in the triangle $ABE$, we can find that $\frac {|BK|}{|KE|}=\frac{|AB|}{|AE|}=\frac{|AC|}{|AE|}=\frac 21$. Thus, $\frac{|BD|}{|DC|}=\frac 21=\frac {|BK|}{|KE|}\Rightarrow KD//EC$. We know that $ABH'C$ is cyclic. Then, $\angle BH'K=\angle BCA=\angle BDK\Rightarrow BH'DK$ is cyclic. Look at the triangle $BKD$. We have $KH\perp BD$ and the reflection of $H$ wrt $BD$ lies on $(BKD)$. Hence, $H$ is the orthocenter of the triangle $BKD$. Thus, $DH\perp BE$.

Attachments:

I coord bashed this one too, as the isosceles condition made it a bit friendlier to coord bash

My sol:- Let $G$ be centroid of ∆$ABC$.$A-G-H$ are collinear as $AB$=$AC$.$BH$ is perpendicular to $GD$ as $GD$ is parallel to $AC$. So $H$ is also the orthocentre of $GDB$. So, done.

Let G is the intersection of AH and BE $\Rightarrow$ G is the centroid of $\triangle ABC$. Using Thalet theorem, it is easy to prove that H is also the orthecenter of triangle GDB $\Rightarrow$ GH $\perp$ BE

Maybe overkill. Let $M$ be the midpoint of $BC$ and $N$ be the foot of altitude from $C$ to $AB$. Let $(BMHN)$ be $\omega$ and $(CHA)$ be $\Omega$. Note that from the isosceles condition, $BC$ is tangent to $\Omega$ at $C$. Let $K=\omega\cap\Omega$. It is well known that $K$ is the $B-$humpty point of $\triangle ABC$, which lies on $BE$. Since $\angle BKH$ is a right angle therefore it is sufficient to prove $K,H,D$ are collinear. Consider $f:R^2\to R$ such that $f(\bullet)=P(\bullet,\omega)-P(\bullet,\Omega)$. It is known that $f$ is linear. We want $f(D)=0$. Now \begin{align*} f(D)&=\frac{1}{3}f(B)+\frac{2}{3}f(C)\\ &=\frac{f(B)+2f(C)}{3}\\ &=\frac{(2\cdot CM\cdot CB-0)+(0-BC^2)}{3}\\ &=\frac{BC^2-BC^2}{3}\\ &=0 \end{align*}So we are done.

Let $B(0,0) ,C(p,0) ,A(\frac{p}{2},n)$ line AH meet BC at midpoint thus $H(\frac{p}{2},Y*)$ we need to find $Y*$ grad $AB$ =$\frac{2n}{p}$ tangent to $AB = \frac{-p}{2n}$ line equation of projection C to AB $y-0=\frac{-p}{2n}(x-p)$ since the line equation will pass through orthocenter we need to find when $x = \frac{p}{2}$ $Y* = \frac{-p}{2n}(\frac{p}{2}-p)=\frac{p^2}{4n}$ $Y* = \frac{p^2}{4n}$ $E = (\frac{3p}{4},\frac{n}{2})$ (midpoint theorem in analytic) $D=(\frac{2p}{3},0)$ thus we need to show only $grad (BE)*grad(HD)=-1$ $\frac{4n}{6p}*\frac{-6p}{4n}=-1$ which shows $BE \perp HD$

Let $AF \perp BC $ such that $F \in BC$ We assume that the perpendicular from $D$ to $BE$ intersects $AF$ at $K$. $H'$ is the orthogonal on $BDK$. $\implies BH' \perp KD$ Let's prove that $KD \parallel EC$. By Menelaus $\frac{BF}{BC}.\frac{CE}{EA}.\frac{AK}{KF}=1$ $\implies \frac{FK}{KA}=\frac{1}{2}=\frac{FD}{DC}$ $\implies KD \parallel AC$

Bashing with complex numbers. let $x$ be the complex associated with the point $X$ We make a homothety so that the circumradius of $\odot ABC $ is 1, now we rotate the graph until $A$ coincides with point 1, then $AO$ is the real line. So $o=0$ Let $M:\text{ midpoint of } BC$ Claim 1: $1=bc$ Proof: Since $AB=AC, \Rightarrow M\in \mathbb{R}$, so $m=\frac{b+c}{2}$ M,O,A are collinear $\Rightarrow \frac{a-o}{\overline{a}- \overline{o}}=\frac{a-m}{\overline{a}-\overline{m}}$ \[2-\frac{bc}{b+c}=2-\frac{1}{b+c}\Rightarrow bc=1 \square\] So, by claim 1 $\overline{b}=c,\overline{c}=b$ Now $d=\frac{2c+b}{3},\overline{d}=\frac{2b+c}{3}$, $e=\frac{a+c}{2}$ Finishing $BE \perp HD\iff \frac{b-e}{h-d}\in\mathbb{R}_i$ \[\iff\frac{3(2b-c-1)}{2(2b+3+c)}\in \mathbb{R}_i\]\[\iff 2b^2+5b-2c^2-5c=-(5c-2c^2+5b+2b^2)\square\] As desired.

Another proof by complex bash same as above let a=vw and b=v² , c=w² . Then d=2w²+v²/3 . Then prove that (h-d)/(b-e) is purely imaginary. In this proof terms get neetly factorised