A geometry problem asking to prove point of intersection of $2$ cevians lies on circumcircle of the triangle.

Question

The points $P$ and $Q$ are chosen on the side $BC$ of an $acute$$angled$- $triangleABC$ so that $\angle PAB=\angle ACB$ and $\angle QAC=\angle CBA$.
The points $M$ and $N$ are taken on the rays $AP$ and $AQ$, respectively, so that $AP=PM$ and $AQ=QN$. Prove that the lines $BM$ and $CN$ intersect on the circumcircle of the triangle ABC.

I tried this problem and found that,

$\angle AQC=\pi-\angle QAC-\angle ACQ$ & $\angle APB=\pi-\angle ABP-\angle PAB$
$\Rightarrow$ $\angle AQC=\angle APB$ ----(1)
$\Rightarrow$$AQ=QN=AP=PM$ & $BC\parallel MN$ -{by midpoint theorem and eq. (1)}
Also that $\triangle ABC\sim \triangle QAC \sim \triangle PBA$

I can't do after this.
Sorry for not providing with figure.
PS: Please give only analytic proofs.

Answer 1

[As pointed out, the following proof is incorrect. See below for the correct version given by @maxim.]

According to your finding:- (1) $AQ = … = AP$; and (2) $\triangle QAC \sim \triangle PBA$,

we can say that $\triangle QAC \cong \triangle PBA$. This means $AC = AB$. In other words, $\triangle ABC$ is isosceles. Then, the figure must be re-drawn as below:-

After forming the red circle (centered at P, radius = AP, diameter = AM), we have $\angle AYM = 90^0$. Similarly, $\angle AYN = 90^0$. Also, $\angle AZB = 90^0$.

Since $90^0 = t + x_1 = t + x_2 = t + x_3$, X is another end point of the diameter of the circle ABC. Result follows by observing that CN will cross AY also at X by symmetry.

Answer 2

From $\triangle ABP \sim \triangle CAQ$, we have $AP/BP = CQ/AQ$, therefore $MP/BP = CQ/NQ$, and $\triangle MBP \sim \triangle CNQ$ by side-angle-side. Then, from $\triangle ABM$, $\beta + \gamma + \delta + \epsilon = \pi$, and $ABRC$ is cyclic.

$\triangle ABC$ does not have to be acute-angled.