BdMO Online Forum

Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$.

Proposed by Giorgi Arabidze, Georgia.

Let $$BM \cap CN=\chi$$.
As $$\angle PAB = \angle BCA $$ and $$\angle CAB = \angle ABC$$,
$$\triangle{ABP} \sim \triangle {ACQ}$$.
$$\Rightarrow \angle{APB}=\angle{AQC}= \angle{A}$$.
And $$\dfrac{AP}{BP}\dfrac{CQ}{AQ}$$
$$\Rightarrow \dfrac{PM}{BP}= \dfrac{CQ}{AN}$$.
So, $$\triangle{BPM}~ \triangle{CQN}$$.
$$\therefore \angle{PBM}=\angle{CNQ}$$ and $$\angle{BMP}= \angle{NCQ}$$.
Now, $$AP=PM$$, $$AQ=AN$$.
$$\therefore PQ \parallel MN \parallel BC$$
$$\Rightarrow \angle{ANM}= \angle{AMN}= \angle{A} $$.
And $$\angle{NCQ}=\angle{CNM}$$, $$\angle{PBM}=\angle{BMN}$$.
Now, $$\angle{BMN} + \angle{CNM}=\angle{BMN} + \angle{BMP}=\angle{PMN}=\angle{A}$$
$$\angle{B\chi C}+\angle A=\angle{M\chi N}+\angle A=\angle180- \angle A+ \angle A=180$$.
$$\therefore A,B,\chi,C$$ are concyclic. [where $$\chi$$ is the intersection of $$BM \text{ and } CN$$]