areacirclescontest-matheuclidean-geometrygeometry
$A,B,C$ and $D$ are concyclic . $AC$ is the diameter of the circle and $AD=DC$ . The area of quadrilateral $ABCD$ is $20c$m$^2$.
Draw a line $DE$ such that $E$ is a point on $AB$, and $DE$ $\bot$ $AB$.
Find the length of $DE$.
The answer keys says that it is ${2}\sqrt{5}$ $cm$. How was the length found?
I imagine that the use of similar triangles and using the fact that angle in a semicircle is $90^\circ $may be involved, but I cannot go beyond that.
Let $\mathcal{C}$ be a circle wichi is tangent to $BC,CD$ and $DA$ and let it center be $O$. Then $O$ must lie on angle bisector of angle $\angle ADC$ and angle $\angle DCB$. So these angle bisectors meet on $AB$.
Now let $E$ be on $AB$ so that $AE=AD$. We have to prove $BC=BE$ i.e. triangle $BCE$ is isosceles. 
If $\angle ADE =\alpha$ then $\angle AED =\alpha$ and $\angle ADE =180^{\circ}-2\alpha$. So $\angle BCD =2\alpha$ and thus $\angle OCD =\alpha$. Since $\angle DEB =180^{\circ}-\alpha$ we have $CDEO$ is cyclic.
Let $\angle OEC =\beta$ then $\angle ODC =\beta$ and $\angle ADO = \beta $ so $\angle ABC =180^{\circ}-2\beta$ and thus $\angle BCE =\beta$.
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.
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