Consider semicircle with diameter \(AB\), \(C\) is a point, \(D\) is the midpoint of arc \(BC\). \(AD\) meets \(BC\) at \(E\). If \(CE=3\), and \(BD=2\sqrt5\), find \(AB\).
Solution
Answer: \(10\)
Extend \(AC\) and \(BD\) to meet at \(P\). Since \(D\) is the midpoint of arc \(BC\), \(DP=BD=2\sqrt5\). The points \(C,E,D,P\) are concyclic.
Using the power of point \(B\),
\[BE\cdot BC=BD\cdot BP=(2\sqrt5)(4\sqrt5)=40.\]
Since \(BC=BE+3\), this gives \(BE=5\) and \(BC=8\). The construction gives \(CP=4\), and the solution concludes \(AB=10\).