Someone gave me this question, and I seriously have no clue how to answer it. I've tried considering the centre, drawing diagrams, and searching up all the theorems in the internet but I can't find a way to solve it. Any help would be greatly appreciated.
There is a pentagon $ABCDE$ inscribed in a circle, with $AB = 2, BC = 5, CD = 2, DE = 5$ and $AD = 8$. What is the length of the line segment $BE$?
It feels like there isn't enough information. I don't even think we can find or use the radius of the circle. There must be a trick somewhere, right?
Best Answer
Let $X$ be the intersection of segments $AD$ and $BE$. From the repeating values in the lengths given, $ABCD$ and $BCDE$ are trapezoids, so $BCDX$ is a parallelogram, which means $|DX|=5$, $|AX|=8-5=3$, and $|BX|=2$. By the intersecting chords theorem, $|XE|={15\over2}$, whence $|BE|={19\over2}$.