Japan Mathematical Olympiad

$4\sqrt{7}$ Let $M$ be the midpoint of segment $BC$, and let $Q$ (resp. $R$) be the intersection of the line through $P$ parallel to segment $BC$ with segments $AB$ (resp. $AC$). As $AB=AC$, we have $\angle ABM=\angle AQP=\angle ARP$, which yields that the right-angled triangles $AMB$, $PFQ$, $PER$ are similar. Thus $PQ:PR=PF:PE=5:2$. Also, the quadrilateral $BCRQ$ is an isosceles trapezoid and those bases are perpendicular to the segment $PD$, so that $PQ-PR=BD-CD=4$. Therefore, $PQ=4\cdot \frac{5}{5-2}=\frac{20}{3}$, and so we have $AB:AM=PQ:PF=4:3$.

Since $AB:BM=AB:\sqrt{A{B}^2-A{M}^2}=4:\sqrt{4^2-3^2}=4:\sqrt{7}$ by the Pythagorean theorem, we have $AB=BM\cdot \frac{AB}{BM}=\frac{9+5}{2}\cdot \frac{4}{\sqrt{7}}=4\sqrt{7}$.