Calculating last two angles of a quadrilateral when two angles and (relative) side lengths are known. I've recently run into a problem that I can't seem to get. Given a quadrilateral, $ABCD$, with angle $D$ equaling 54 deg, and with the length of $BC$ equaling $CD$ while being double the length of $BA$, find the measure of angle $C$ in degrees. The last piece of given information is that $BA$ and $AD$ are perpendicular.

 A: draw a perpendicular $CE$from $C$ to $AD$ so that the foot $E$ is on $AD.$ draw another perpendicular $BF$ from $B$ to $CE.$  we will compute $EC$ in two ways: $$EC = x \sin 54^\circ = CF + FE = CF + AB = x\sin\angle CBF + \frac x 2   $$  therefore $$\sin \angle CBF=\frac{2\sin 54^\circ - 1}{2} \to \angle CBF = 53^\circ $$
therefore $$\angle BCD = 360^\circ-(\angle A + \angle B + \angle D) =360^\circ-(90^\circ+ 90^\circ + 53^\circ + 54^\circ)=73^\circ  $$
A: Follow the steps,same as above. The only mistake is in the calculation of angle CBF . Its 18 degrees.
So in the final step 
angle BCD is 108 degrees
A: An alternative way of doing this problem is to consider the reflection of the quadrilateral $ABCD$ in the line $AD$ so that the combined figure is a five-sided polygon (only 5 sides since $BA$ is perpendicular to $DA$). Now the pentagon has one angle of $108$ but now all sides equal $x$. Therefore it is a regular pentagon and therefore $\theta=108$.
The essence of the problem is that the original quadrilateral is half a regular pentagon.
