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.

  • $\begingroup$ 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) $\endgroup$ – SmarthBansal May 10 '18 at 18:37
  • $\begingroup$ SmarthBansal- Yes, but I don't get there. $\endgroup$ – Love Invariants May 10 '18 at 18:39
  • $\begingroup$ What's $AD$, $DN$? Do you mean $AQ=AN$? $\endgroup$ – SmarthBansal May 10 '18 at 18:42
  • $\begingroup$ SmarthBansal- yes. $\endgroup$ – Love Invariants May 10 '18 at 18:43
  • $\begingroup$ Don't abuse $…$ notation for italic text (as opposed to formula symbols). Use *…* for that instead. $\endgroup$ – 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:-

enter image description here

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.

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

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

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