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$ABC$ 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

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  • $\begingroup$ Have you seen a figure and written down everything you are told on that figure? That usually helps a bit, at least with getting some overview over all the knowns and unknowns. $\endgroup$ – Arthur Dec 15 '17 at 6:48
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$$S_{\Delta APQ}=\frac{1}{2}\cdot6\cdot20\sin\measuredangle A$$ 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 $x=4.$

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  • $\begingroup$ thanks i was using angle b did'nt think about a $\endgroup$ – 3.14159 Dec 15 '17 at 9:16
  • $\begingroup$ @3.14159 You are welcome! $\endgroup$ – Michael Rozenberg Dec 15 '17 at 9:17

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