$ABC$ is a triangle. $P$ and $Q$ are points on $AB$ and $AC$ of the triangle. $AP=6,$ $AQ=20.$ Area of triangle $APQ$ is equal to area of quadrilateral $PQBC.$
If $PB=QC$ what is length of $PB$ given $BC =25$. Given BC =25 Find $x$ if $PB=QC=x$ Im getting too many equations with too many unknowns. Any insight into problem will help
$$S_{\Delta APQ}=\frac{1}{2}\cdot6\cdot20\sin\measuredangle A$$ and $$S_{\Delta ABC}=\frac{1}{2}\cdot(6+x)(20+x)\sin\measuredangle A.$$ Thus, since $$S_{\Delta ABC}=2S_{\Delta APQ},$$ we obtain $$(6+x)(20+x)=2\cdot6\cdot20,$$ which gives $x=4.$