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).
Best Answer
Reflect $A$ across $D$ to a new point $Q$. Then $\triangle QDE\cong \triangle ECB$ (sas).
Since $$\angle QEC = \angle DEB $$ we see that quadrilateral $ACEQ$ is cyclic, so $$\angle QCE =\angle QAE = \angle DAE = \angle DNE$$ so $QC||DN$. Now since $D$ halves $AQ$ the line $DI$ is a middle line for triangle $QAC$ and thus it halves $AC$.