Geometry – Finding the area of quadrilateral PC1QC2

Question

$ABCD$ is a quadrilateral with side lengths $AB=4$, $BC=10$, $CD=6$ and $AD=6$, and diagonal $BD=8$. If the incircles of triangles $ABD$ and $BCD$ touch $BD$ at $P$ and $Q$ respectively, then area of the quadrilateral $C_1PC_2Q$ (where $C_1$ and $C_2$ are the incenters of the $\triangle ABD$ and $\triangle BCD$ respectively), is?

Answer 1

You found $DQ=2$ correctly. Now you must find $BP$. Apply Cosine theorem for $\triangle ABD$: $$AD^2=AB^2+BD^2-2AB\cdot BD\cdot \cos ABD \Rightarrow \\ \cos ABD=\frac{4^2+8^2-6^2}{2\cdot 4\cdot 8}=\frac{11}{16} \Rightarrow \\ \cos (2\cdot \frac{ABD}2)=2\cos^2(\frac{ABD}2)-1=\frac{11}{16} \Rightarrow \\ \cos^2(\frac{ABD}2)=\frac{27}{32}=\frac{1}{1+\tan^2 (\frac{ABD}2)}\Rightarrow \\ \tan (\frac{ABD}2)=\frac{\sqrt{5}}{3\sqrt{3}}=\frac{C_1P}{BP}\Rightarrow \\ BP=\frac{3\sqrt{3}\cdot \frac{\sqrt{15}}{3}}{\sqrt{5}}=3$$ Hence, $PQ=BD-BP-DQ=8-3-2=3$.

Can you finish it yourself?

Answer:

$A=\frac12\cdot PQ\cdot\left(C_1P+C_2Q\right)=\frac12\cdot 3\cdot \left(\frac{\sqrt{15}}{3}+2\right).$