The line $DN$ bisects the line segment $AC$ if $AD=BC$. Consider a circle with diameter $AB$.  A point $D$ on the circle is chosen arbitrarily such that $D\ne A,B$.   A point $C\in AB$ is also chosen arbitrarily such that $C\ne A,B$. Draw $CH$ perpendicular to $AD$ at $H$. The internal angular bisector of $\angle DAB$ intersects the circle at $E$, and intersects $CH$ at $F$.  The line $DF$ intersects circle again at $N$. 

Prove that the points $N$, $C$, and $E$  are collinear.  If $AD=BC$, then prove that $DN$ intersects $AB$ at the midpoint $I$ of $AC$.


I proved $N,C,E$  are collinear by cyclic quadrilateral and two congruent triangles. But I have no idea for the second problem.

 A: 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 $AC$. 
A: This is a trigonometric solution, so it is not very nice. 
 In this part of my answer, I shall prove the first part of the problem.
 
Let $\theta:=\angle ABD$, $\alpha:=\angle HFD$, and $\beta:=\angle BEC$.  Write $d$ for the diameter of the circle (i.e., $d=AB$), and $l$ for the length of $BC$.  We see that
$$CH\parallel BD\text{ so that }DH=l\sin(\theta)\,.$$
Now, $\angle DAE=\angle EAB=\dfrac{\pi}{4}-\dfrac{\theta}{2}$.  Hence,
$$\angle AFD=\frac{\pi}{4}+\frac{\theta}{2}+\alpha\,.$$
Since $AD=d\sin(\theta)$, we get
$$\frac{DF}{\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\frac{d\sin(\theta)}{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)}$$
via the Law of Sines on the triangle $ADF$.  Thus,
$$\sin(\alpha)=\frac{DH}{DF}=\frac{l\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)}{d\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\,.\tag{1}$$
Note that $\angle AEB=\dfrac{\pi}{2}$, so
$$BE=d\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\,.$$
Applying the Law of Sines on the triangle $BEC$ yields
$$\frac{l}{\sin(\beta)}=\frac{d\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\beta\right)}\,,\tag{2}$$
noting that $\angle EBD=\angle EAD=\dfrac{\pi}{4}-\dfrac{\theta}{2}$.  Consequently,
$$\frac{\sin(\beta)}{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\beta\right)}=\frac{l}{d\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=\frac{\sin(\alpha)}{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)}\,,$$
using (1) and (2).  Using the trigonometric identity $\sin(x)\sin(y)=\dfrac{1}{2}\big(\cos(x-y)-\cos(x+y)\big)$, we obtain
$$\cos\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha-\beta\right)=\cos\left(\frac{\pi}{4}+\frac{\theta}{2}-\alpha+\beta\right)\,.$$
Hence,
$$\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\sin(\alpha-\beta)=0\,.$$
Because $0<\theta<\dfrac{\pi}{2}$, we get $\sin\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)>0$, making $\sin(\alpha-\beta)=0$.  As $\alpha$ and $\beta$ belong to the interval $\left(0,\dfrac{\pi}{2}\right)$, the condition $\sin(\alpha-\beta)=0$ implies that $\alpha=\beta$.
Finally. we see that $\angle AEB=\dfrac{\pi}{2}$, so
$$\angle AEC=\frac{\pi}{2}-\beta=\frac{\pi}{2}-\alpha=\angle ADN=\angle AEN\,.$$
This shows that $E$, $C$, and $N$ are collinear.

I shall now prove the second part of the problem.  In fact, I shall prove a stronger result that $DN$ meets $AB$ at the midpoint $I$ of $AC$ if and only if $AD=BC$.
Let $G$ be the point of intersection between $DN$ and $AB$.  Using the Law of Sines on the triangle $GFC$ noting that $\angle GFC=\alpha$ and $\angle GCF=\theta$, we get
$$GC=GF\left(\frac{\sin(\alpha)}{\sin(\theta)}\right)\,.$$
We have
$$\frac{GF}{GD}=\frac{AG}{AG+AD}=\frac{AG}{AG+d\sin(\theta)}$$
by the Angle Bisector Theorem.  Therefore, $G=I$ if and only if $AG=GC$, which is equivalent to
$$\frac{\sin(\theta)}{\sin(\alpha)}=\frac{GD}{AG+d\sin(\theta)}\,.$$
The Law of Sines on the triangle $AGD$ gives
$$AG=AD\left(\frac{\cos(\alpha)}{\sin(\alpha+\theta)}\right)=\frac{d\sin(\theta)\cos(\alpha)}{\sin(\alpha+\theta)}$$
and
$$GD=AD\left(\frac{\cos(\theta)}{\sin(\alpha+\theta)}\right)=\frac{d\sin(\theta)\cos(\theta)}{\sin(\alpha+\theta)}\,.$$
That is, $G=I$ iff
$$\frac{\sin(\theta)}{\sin(\alpha)}=\frac{\cos(\theta)}{\cos(\alpha)+\sin(\alpha+\theta)}\,.\tag{3}$$
Note that 
$$\begin{align}
\cos(\alpha)+\sin(\alpha+\theta)&=\sin\left(\frac{\pi}{2}-\alpha\right)+\sin(\alpha+\theta)
\\&=2\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}-\alpha\right)
\\&=2\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)\,,
\end{align}$$
where we have implemented the identity $$\sin(x)+\sin(y)=2\sin\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right)\,.$$
Furthermore, from $\sin(2x)=2\sin(x)\cos(x)$, we have
$$\begin{align}\cos(\theta)&=\sin\left(\frac{\pi}{2}-\theta\right)\\&=2\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\,.\end{align}$$
That is, (3) is equivalent to
$$\sin(\alpha)=\sin(\theta)\left(\frac{\sin\left(\frac{\pi}{4}+\frac{\theta}{2}+\alpha\right)}{\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\right)\,.$$
Combining the last result with (1), we conclude that $G=I$ if and only if $$BC=l=d\sin(\theta)=AD\,.$$ 
