Dear Mathlinkers, a similar one https://www.facebook.com/photo.php?fbid=10215351043824082&set=g.1019808738132832&type=1&theater&ifg=1 Sincerely Jean-Louis

this problem can be done using pop+stewart's theorem+angle bisector theorem, or using the distance formula in barycentric coordinates. I can post my solution if anyone wants me to do so.

Let $BC = a, CA = b, AB = c$. It is known $Q$ is arc midpoint of $BC$ in $(ABC)$. $\angle ABP = \angle ABC = \angle AQC = \angle PQC = \angle CQP$ $\angle APB = \angle CPQ$ Thus $\triangle ABP \sim \triangle CQP$, and $\frac{AP}{BP} = \frac{CP}{QP}$ $PQ = \frac{BP \cdot CP}{AP}$ By Angle Bisector Theorem, $BP = \frac{ac}{b+c}$ and $CP = \frac{ab}{b+c}$. So $PQ = \frac{BP \cdot CP}{AP} = \frac{\frac{a^2bc}{(b+c)^2}}{AP}$ $\angle BAQ = \angle PAC$ and $\angle AQB = \angle ACB = \angle ACP$ so $\triangle ABQ \sim \triangle APC$. $\frac{AB}{AQ} = \frac{AP}{AC}$ $AQ = \frac{AB \cdot AC}{AP} = \frac{bc}{AP}$ $\frac{PQ}{AQ} = \frac{a^2bc}{bc(b+c)^2} = \frac{a^2}{(b+c)^2} = \left( \frac{BC}{AB + AC}\right)^2$.

So obviously we have that $Q$ lies on the circle around $\triangle ABC$. So I shall start off with a simple angle-chase: $$ \angle BAQ = \angle BCQ = \frac{1}{2} \angle A$$Of course the quad $BACQ$ is cyclic, thus Ptolemy's theorem holds, and $AQ$ can be expressed as: $$AQ = BQ.\frac{AC+AB}{BC} $$Now onto some trigonomtry: From the Angle-Bisector theorem we have that $PC=\frac{AC}{AB} BP$. With this in mind,let's apply the sine law on the $\triangle PQC$ and on $\triangle BPQ$. Now we get the following results (keep in mind $\angle QPC = \angle C + \frac{1}{2} \angle A$): $$ PQ=\frac{sin \frac{1}{2} \angle A}{sin(\angle C + \frac{1}{2} \angle A)} BQ$$Now we have the following when we input this into our ratio ($\frac{PQ}{AQ}$): $$\frac{PQ}{AQ} = \frac{sin \frac{1}{2} \angle A}{sin(\angle C + \frac{1}{2} \angle A)}.\frac{BC}{AC+AB}$$Thus we need to show that $\frac{sin \frac{1}{2} \angle A}{sin(\angle C + \frac{1}{2} \angle A)} = \frac{BC}{AC+AB}$ But we applying the sine rule onto the $\triangle ABC$ and representing $\angle B = 180 - \angle A - \angle C$.This statement get's boiled down to the following: $$ 1 + cos \angle A = 2cos^2 \frac{1}{2} \angle A $$Which is an identity, thus the statement of the problem is true.....

Easy be Stewart's Theorem

Note $Q$ lies on the circumcircle of $\triangle ABC$. By Ptolemy's theorem on $ABQC$ and since $QB=QC$, \[AQ\cdot BC=(AB+AC)QB\;\Longrightarrow\;\frac{QB}{AQ}=\frac{BC}{AB+AC}.\]$QA\cdot QP=QB^2$ by Shooting lemma (or $\triangle QAB\sim\triangle QBP$): we have \[\left(\frac{BC}{AB+AC}\right)^2=\frac{QA\cdot QP}{AQ^2}=\frac{PQ}{AQ}.\]$\square$

Post for storage By Angle Bisector Theorem: $\frac{PB}{PC} =\frac{AB}{AC} \Rightarrow \frac{PB}{AB} =\frac{PC}{AC} =\frac{BC}{AB+AC}$ We'll prove: $\frac{PQ}{AQ} =(\frac{BC}{AB+AC})^2$ but $(\frac{BC}{AB+AC})^2 =\frac{PB.PC}{AB.AC} =\frac{AP.PQ}{AB.AC}$ since $ABQC$ is cyclic So it's equivalent to: $AP.AQ=AB.AC$, which can easy obtain through a $\sqrt{bc}$ inversion

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/78.%200.%20Relations%20metriques.pdf (Problem 28) Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, here (Problem 28) Sincerely Jean-Louis