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I have this problem:

enter image description here

My attempt was:

Since, $\angle AFB = \angle ABF => AF = AB = FC$

And $\triangle AFB$ isosceles of base $FB$

According to bisector theorem $CQ = 2BQ$, then point $Q$ is not midpoint, therefore point $P$ is not the barycentre of $\triangle ABC$

and I could not get more information. Any hint is appreciated.

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  • $\begingroup$ Constructing this in Geogebra, PQ=3, if that inspires anyone to a proof. $\endgroup$ – Matthew Daly Aug 21 '19 at 22:32
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enter image description here

Another solution:

Since $AP$ is the angle bisector and is also perpendicular to $BF$, it follows that $AB = AF$. By the angle bisector theorem, it follows that $$ \frac{BQ}{CQ} = \frac{AB}{AC} = \frac{1}{2}. $$ Now, noticing that $\triangle ABQ $ and $\triangle AFQ$ share an edge $AQ$, $AB = AF$, and $\angle BAQ = \angle FAQ$, it follows that the two triangles are congruent and hence $BQ = FQ$. Let $M$ be the midpoint of $CQ$. Then, we have $$BQ = QM = MC = QF.$$ This shows that $\triangle BFM$ is a right triangle with $\angle BFM = 90^\circ$. Hence, we also have $AQ \parallel FM$. Set $PQ = x$. Then, $$ \frac{x}{FM} = \frac{PQ}{FM} = \frac{1}{2} = \frac{FM}{AQ} = \frac{FM}{9+x} $$ Noting that $FM = 2PQ = 2x$ yields $$ 4x^2 = x(9+x) \implies x = 3 \implies PQ = 3 $$ as desired.

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By Apollonius's theorem, we have $$BA^2 + BC^2 = 2BF^2 + 2AF^2$$ Subtracting $BA^2$ from both sides yields $$BC^2 = 9BQ^2 = 2BF^2 + AF^2 = 8BP^2 + AF^2$$ Also, $BQ^2 = BP^2 + PQ^2$ so $$9BQ^2 = 9BP^2 + 9PQ^2 = 8BP^2 + AF^2$$ iff $BP^2 + 9PQ^2 = AF^2$ iff $$9PQ^2 = AF^2 - BP^2 = AF^2 - PF^2 = AP^2$$ So $9PQ^2 = AP^2 = 81$ which means $PQ = 3$.

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