# How to get the measure?

I have this problem:

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.

• Constructing this in Geogebra, PQ=3, if that inspires anyone to a proof. – Matthew Daly Aug 21 '19 at 22:32

## 2 Answers

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.

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 $$PQ = 3$$.