0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

[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.

$\endgroup$
  • $\begingroup$ @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
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.