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

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.

• Proving sum of opposite angles of $ABFC$ are $180^\circ$ should do the trick. (Where F is the point of intersection of BM and NC) – SmarthBansal May 10 '18 at 18:37
• SmarthBansal- Yes, but I don't get there. – Love Invariants May 10 '18 at 18:39
• What's $AD$, $DN$? Do you mean $AQ=AN$? – SmarthBansal May 10 '18 at 18:42
• SmarthBansal- yes. – Love Invariants May 10 '18 at 18:43
• Don't abuse $…$ notation for italic text (as opposed to formula symbols). Use *…* for that instead. – MvG May 10 '18 at 19:50

[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.

• @Mick- Thank you. – Love Invariants May 13 '18 at 13:47
• @LoveInvariants you are welcome. – Mick May 13 '18 at 13:56
• It's not true that $AB$ has to be equal to $AC$. $\triangle QAC \sim \triangle PBA$ and $AP = AQ$ doesn't imply that. – Maxim Jun 8 '18 at 13:52
• @Maxim I must have overlooked the problem. Will leave it there for the moment. – Mick Jun 8 '18 at 19:07

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.

• Your version saves me time in rectifying mine. – Mick Jun 16 '18 at 8:14