In the following problem, $\angle DAB = 2\alpha$, and $ABCD$ is a kite ($AD=AB, DC=CB$).
I need to prove the ratio between the circle inscribed by the kite to the area of the circle inscribing the kite is $\frac{\sin^22\alpha}{1+\sin2\alpha}$.
I proved $\angle ADC = \angle ABC = 90^\circ$, and that $AC$ is the diameter of the circle inscribing the kite. I have difficulties with finding the area of the circle inscribed by the kite, and hope one of you will be able to help me.
Sorry if my English is bad, it is not my native language.
Best Answer
Draw the segments from the center of the inscribed circle to the vertices of the polygon, forming triangles. Each triangle has one side of the quadrilateral as its base and the radius of the circle as the corresponding altitude. Thereby we have
$\text{Radius of inscribed circle}=2×\text{Area of polygon}/\text{Perimeter of polygon}$
In this case the area of the polygon, as you can see is given by two right triangles having the sides of the polygon as legs, and the perimeter is the sum of the legs of both right triangles. That plus what you already know should get you to the radius of the inscribed circle and then you get its area.