First, since the angles $\widehat{A H D}$ and $\widehat{A E D}$ are right angles, the points $A, H, E, D$ lie on the circle with diameter $[A D]$.
Let $\theta=\widehat{C B A}=\widehat{A C B}$. Since $A H C$ is a right triangle at $H$, $\widehat{H A C}=90^{\circ}-\theta$. Since $B A D$ is a right triangle at $A$, $\widehat{A D B}=90^{\circ}-\theta$, and thus $\widehat{A D H}=\widehat{H A E}$.
By the cocyclicity of $A, H, E, D$, the angles $\widehat{A D H}$ and $\widehat{A E H}$ are equal or supplementary, so $\widehat{H A E}=\widehat{A E H}$ or $\widehat{H A E}+\widehat{A E H}=180^{\circ}$.
The second case cannot occur because the sum of the angles in triangle $A E H$ is $180^{\circ}$, so we have $\widehat{H A E}=\widehat{A E H}$. This implies that $H A E$ is isosceles at $H$, hence $H A=H E$.
Another approach using angles between lines. We know that if $T$ is a tangent at a point $A$ to a circle $(C)$ and if $B$ and $M$ are two other points on $(C)$, then $(T, A B)=(M A, M B)$.
In the problem, since $(A B)$ is perpendicular to the diameter $(A D)$, it is tangent to the circle. Therefore, from the above, we have $(A B, A H)=(E A, E H)$. Since $A B C$ is isosceles, we have $(A B, A H)=(A H, A C)$, so $(A H, A E)=(A H, A C)=(A B, A H)=(E A, E H)$. We conclude that $H A E$ is isosceles at $H$, hence $H A=H E$.