$\triangle$$ABC$ is isosceles triangle that is why $\angle$$DCB$$=$$\angle$$BCE$$=$$\alpha$ Let be $\angle$$ACD$$=2$$\beta$ $\angle$$ICD$$=$$\angle$$ICA$$-$$\angle$$DCA$$=$$\alpha+\beta-$$2$$\beta$$=$$\alpha$$-$$\beta$ Then $\angle$$ICJ$$=$$\angle$$JCD$$-$$\angle$$ICD$$=$ $90-$$\beta$$-(\alpha-\beta)$$=90-\alpha$ in the other hand $\angle$$CIJ$$=$$180-$$\angle$$AIC$$=$ $180-(90+\frac{90-\alpha}{2})=45+\frac{\alpha}{2}$ from there we get $|IC|=|CJ|$ Let $K$ be midpoint of $IJ$.İf we show that $|KB|=|KC|$ then we are done.$\angle$$CKA$$=90$ and $\angle$$ABC$$=90$ then we get $ACKB$ is cyclic.And we know $AK$ is $\angle$$A$ bisector.That is why we get $|KB|=|KC|$.we are done.

Here's a not-so-nice solution. We may completely eliminate point $B$ from the picture, restating the problem: Quote: Let $AEC$ be a triangle and $D$ a point on $AE$ such that $CD=CE$. Let $I$ be the incenter of triangle $ACE$ and $J$ the $A-$excenter of triangle $ADC$. Lastly, let $M$ be the midpoint of $EC$ and $N$ be the midpoint of $IJ$. Then, $MN$ and $AE$ are parallel. Let $AE=x,AC=y,EC=z$. Let $AI$ intersect $EC$ at point $Q$. Moreover, let $T,K,S,L$ be the feet of the perpendiculars from points $C,I,Q,J$ to $AE$, respectively. It suffices to prove that $\dfrac{AQ}{QN}=\dfrac{EQ}{QM}$. We split the remaining of the proof into two parts: Part 1: Computing $\dfrac{EQ}{QM}$. Note that $\dfrac{EQ}{QC}=\dfrac{EA}{AC},$ and so $EQ=\dfrac{xz}{x+y},$ therefore $QM=EM-EQ=\dfrac{z(y-x)}{2(x+y)}$. Thus, $\dfrac{EQ}{QM}=\dfrac{2x}{y-x}$. Part 2: Computing $\dfrac{AQ}{QN}$. Note that $2QN=2(IN-IQ)=2IN-2IQ=IJ-2IQ=IJ-2(AQ-AI)=$ $=IJ+2AI-2AQ=(AI+AJ)-2AQ$, and so $\dfrac{AQ}{QN}=\dfrac{2AQ}{AI+AJ-2AQ},$ therefore we equivalently have to compute $\dfrac{AI+AJ}{AQ}$. Now, note that $\cos \dfrac{\angle A}{2}=\dfrac{AK}{AI},$ hence $AI=\dfrac{x+y-z}{2\cos \dfrac{\angle A}{2}}$. Similarly, $AJ=\dfrac{AL}{\cos \dfrac{\angle A}{2}}=\dfrac{AD+y+z}{2\cos \dfrac{\angle A}{2}}=\dfrac{\dfrac{y^2-z^2}{x}+y+z}{2\cos \dfrac{\angle A}{2}},$ with the last equality being true as $AD=TA-TD=TA-TE=\dfrac{AC^2-CE^2}{AT+TE}=\dfrac{y^2-z^2}{x}$. Lastly, $AQ=\dfrac{AS}{\cos \dfrac{\angle A}{2}}=\dfrac{(x+y)^2-z^2}{2(x+y)\cos \dfrac{\angle A}{2}},$ with the last equality being true since $AS=AE-ES=x-\dfrac{ET \cdot EQ}{z},$ and we may easily compute $ET,EQ$. Now, putting these all together, $\cos \dfrac{\angle A}{2}$ cancels out and we have $\dfrac{AI+AJ}{AQ}=\ldots=\dfrac{x+y}{x}$. Therefore, returning back, $\dfrac{AQ}{QN}=\dfrac{2AQ}{AI+AJ-2AQ}=\dfrac{x}{y-x}$. Thus, the two ratios are equal, as desired.

A nice solution Let $K$ be the intersection of the parallel drawn from point $C$ to $AB$ and the line $AI$.Since $\angle BAK=\angle CAK$ we get that $\angle CAK=\angle CKA$ so we get that $CA=CK$. By easy angle chasing (using the fact I and J incenter and excenter) we get that $\angle CIJ=\angle CJI$ so we get that $CI=CJ$ From $CI=CJ$ and $CA=CK$ we get that triangles $CIA$ and $CJK$ are congruent so $IA=JK$. If we show that the perpendicular bisector of $BC$ passes through the midpoint of $AK$ then we are done (because $IA=JK$). Since $AB\parallel CK$, perpendicular bisector of $BC$ passes through the midpoint of $AK$. We are done