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The problem can be reduced to isosceles trapzoid ==> isosceles triangle AEB ==>equilateral triangle AEB given $\angle AEB = 60^o$. $D$ happens to the be midpoint of $AE$
A quadrilateral $ABCD$ is inscribed in a circle. Suppose that $|DA| =|BC|= 2$ and$ |AB| = 4$. Let $E $ be the intersection point of lines $BC$ and $DA$. Suppose that $\angle AEB = 60^o$ and that $|CD| <|AB|$. Calculate the radius of the circle.
The problem can be reduced to isosceles trapzoid ==> isosceles triangle AEB ==>equilateral triangle AEB given $\angle AEB = 60^o$. $D$ happens to the be midpoint of $AE$