Triangle $ABC$ has side lengths $\overline{AB}=2\sqrt{3}$ and $\overline{BC}=2,$ as shown. $D$ is the midpoint of $\overline{BC},$ satisfying $\overline{AD}=\sqrt{7}.$

Let $E$ be the point of intersection between $\overline{AB}$ and the bisector of $\angle ACB.$ $\overline{CE}$ meets with $\overline{AD}$ at point $P,$ and the bisector of $\angle APE$ meets with $\overline{AB}$ at point $R.$ An extension of $\overline{PR}$ meets with $\overline{BC}$ at point $Q.$

Given that the area of $\triangle PQC$ is $a+b\sqrt{7}$ times larger than that of $\triangle PRE$ for some rational numbers $a$ and $b,$ find the value of $ab.$

*
This problem is a part of
<Grade 10 CSAT Mock Test> series
.
*

The answer is -16.

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$\overline{AC}=2,$ and therefore $\triangle ABC$ is an isosceles triangle, leading to $\overline{BE}=\overline{AE}=\sqrt{3}$ and $\overline{CE}=1.$

Since $\overline{CE}$ and $\overline{AD}$ are both midlines of $\triangle ABC,$ $\overline{PE}=\dfrac{1}{3},$ $\overline{PC}=\dfrac{2}{3},$ $\overline{PA}=\dfrac{2\sqrt{7}}{3},$ and $\overline{PD}=\dfrac{\sqrt{7}}{3}.$ $P$ is the centroid of $\triangle ABC.$

$\overline{QR}$ bisects $\angle APE$ and $\angle CPD,$ so we know that

$\overline{RA}:\overline{RE}=\overline{PA}:\overline{PE}=2\sqrt{7}:1,$ and $\overline{QD}:\overline{QC}=\overline{PD}:\overline{PC}=\sqrt{7}:2.$

Then, $\triangle PRE=\dfrac{1}{2\sqrt{7}+1}\triangle APE,$ and $\triangle PQC=\dfrac{2}{2+\sqrt{7}}\triangle CPD.$

$\triangle APE=\triangle CPD$ from $P$ being the centroid, and therefore,

$\begin{aligned} a+b\sqrt{7} &=\dfrac{\dfrac{2}{2+\sqrt{7}}}{\dfrac{1}{2\sqrt{7}+1}} \\ \\ &=\dfrac{2(2\sqrt{7}+1)}{\sqrt{7}+2} \\ \\ &=\dfrac{2(2\sqrt{7}+1)(\sqrt{7}-2)}{3} \\ \\ &=\dfrac{2(12-3\sqrt{7})}{3} \\ \\ &=8-2\sqrt{7}. \end{aligned}$

$\therefore \, ab=8\cdot(-2)=\boxed{-16}.$