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.
$…$
notation for italic text (as opposed to formula symbols). Use*…*
for that instead. $\endgroup$ – MvG May 10 '18 at 19:50