Find the area of ​the $AFRC$ quadrilateral below.

geometry euclidean-geometry plane-geometry

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If $\angle ABC = \theta, \angle CBF = \angle ABR = \angle DBE = 90^\circ + \theta$

We also have $AB = BF, BC = BR$ and $BD:BE = BF:BC = AB:BR$

So $\triangle ABR \cong \triangle FBC$ and $\triangle ABR \sim \triangle DBE$

That leads to $AR \perp CF$ and $AR = CF = 8 \sqrt2$

Therefore $S_{AFRC} = \frac{1}{2} \cdot 8 \sqrt2 \cdot 8 \sqrt2 = 64$

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