Let $\Delta ABC$ be a right triangle. Prove that $\angle BEH=\angle HCI$. Let $\Delta ABC$ is a right triangle. $D$  is chosen arbitrarily in $AB$,the segment $DH$ is perpendicular to the segment $BC$ at $H$, $E\in AC$ such that $DE=DH$. $I$ is the midpoint of $HE$. Prove that $\angle BEH=\angle HCI$.

Let the intersection point of $CI$ and $BE$ be $X$.
So we need to prove that $\angle HCX=\angle XEH$ or $XECH$ is a cyclic quadrilateral.
Or $\Delta BHX$ and $BEC$ are similar triangles (side-angle-side) $$\Leftrightarrow \frac{BE}{BC}=\frac{BH}{BX}.$$
Then I do not know how to get it, and that idea has not used $DE=DH$.

 A: Introduce $\varepsilon_1=\angle BEH$, $\varepsilon_2=\angle HCI$, $d=DH=DE$, $\gamma=\angle C$, $\delta=\angle HDI=\angle EDI$, $a=BC$. We need to prove that $\varepsilon_1=\varepsilon_2$.

Let us first find the connection between $d$ and $\delta$, they are not independent:
$$BA=BD+DA$$
$$a\sin\gamma=\frac{d}{\cos\gamma}+d\cos(\pi-2\delta-\gamma)=\frac{d}{\cos\gamma}-d\cos(2\delta+\gamma)$$
$$\frac ad=\frac{1-\cos\gamma\cos(2\delta+\gamma)}{\sin\gamma\cos\gamma}\tag{1}$$
Now apply law of sines to triangle $\triangle BEH$:
$$\frac{BH}{\sin\varepsilon_1}=\frac{HE}{\sin(\delta-\varepsilon_1)}$$
$$\frac{d\tan\gamma}{\sin\varepsilon_1}=\frac{2d\sin\delta}{\sin(\delta-\varepsilon_1)}\tag{1'}$$
Solve for $\varepsilon_1$ and you get:
$$\cot\varepsilon_1=2\cot\gamma+\cot\delta\tag{2}$$
Now apply law of sines to triangle $\triangle HCI$:
$$\frac{HI}{\sin\varepsilon_2}=\frac{HC}{\sin(\pi-\delta-\varepsilon_2)}=\frac{BC-BH}{\sin(\delta+\varepsilon_2)}$$
$$\frac{d\sin\beta}{\sin\varepsilon_2}=\frac{a-d\tan\gamma}{\sin(\delta+\varepsilon_2)}\tag{2'}$$
Solve for $\varepsilon_2$ and you get:
$$\cot\varepsilon_2=\frac{\frac ad-\tan\gamma}{\sin^2\delta}-\cot\delta\tag{3}$$
Now replace (1) into (3). The expression has to evolve into (2), assuming that the problem statement is true. In the beginning it looks pretty much hopeless but if you survive the first few lines, you will be quickly rewarded:
$$\cot\varepsilon_2=\frac{\frac{1-\cos\gamma\cos(2\delta+\gamma)}{\sin\gamma\cos\gamma}-\tan\gamma}{\sin^2\delta}-\cot\delta$$
$$\cot\varepsilon_2=\frac{1-\cos\gamma\cos(2\delta+\gamma)-\sin^2\gamma}{\sin^2\delta\sin\gamma\cos\gamma}-\frac{\cos\delta}{\sin\delta}$$
$$\cot\varepsilon_2=\frac{\cos\gamma-\cos(2\delta+\gamma)}{\sin^2\delta\sin\gamma}-\frac{\cos\delta}{\sin\delta}$$
$$\cot\varepsilon_2=\frac{\cos\gamma-\cos(2\delta)\cos\gamma+\sin(2\delta)\sin\gamma-\sin\delta\cos\delta\sin\gamma}{\sin^2\delta\sin\gamma}$$
$$\cot\varepsilon_2=\frac{\cos\gamma-\cos(2\delta)\cos\gamma}{\sin^2\delta\sin\gamma}+\frac{2\sin\delta\cos\delta\sin\gamma-\sin\delta\cos\delta\sin\gamma}{\sin^2\delta\sin\gamma}$$
$$\cot\varepsilon_2=\frac{\cos\gamma-(2\cos^2\delta-1)\cos\gamma}{\sin^2\delta\sin\gamma}+\cot\delta$$
$$\cot\varepsilon_2=\frac{2\cos\gamma(1-\cos^2\delta)}{\sin^2\delta\sin\gamma}+\cot\delta$$
$$\cot\varepsilon_2=2\cot\gamma+\cot\delta$$
$$\cot\varepsilon_2=\cot\varepsilon_1$$
Done.
EDIT: Let me show how I got from (1') to (2)
$$\frac{d\tan\gamma}{\sin\varepsilon_1}=\frac{2d\sin\delta}{\sin(\delta-\varepsilon_1)}\tag{1'}$$
$${\tan\gamma}{\sin(\delta-\varepsilon_1)}={2\sin\delta}{\sin\varepsilon_1}$$
$${\tan\gamma}{(\sin\delta\cos\varepsilon_1-\cos\delta\sin\varepsilon_1)}={2\sin\delta}{\sin\varepsilon_1}$$
Now divide both sides with ${\sin\delta}{\sin\varepsilon_1}{\tan\gamma}$:
$$\cot\varepsilon_1-\cot\delta=2\cot\gamma$$
$$\cot\varepsilon_1=2\cot\gamma+\cot\delta$$
...which is (2). The same procedure will take you from (2') to (3).
