Triangle inscribed in a circle

Question

Isosceles triangle ABC with vertex at A is inscribed in a circle. Chord AD intersects BC at E.

Prove that the square of the length of AB minus the square of the length of AE is equal to the product of the length of BE times the length of CE.

Answer 1

Mark is correct (as he most often is!

I will not use the vinculum .

F is the midpoint of BC

AB² = AF² + BF²

AE² = AF² + EF²

^{Then a little algebra gets the answer!}

Answer 2

I recommend using indirect proof, AE^2 = (BE)(CE).

Then use (BE)(CE) = (AE)(ED) from the intersecting chords.

With substitution, AB^2 - AE^2 = (AE)(ED).

AB^2 = (AE)(ED) + AE^2

=(AE)(ED + AE) = (AE)(AD)

I'm looking for the use of secant-tangent line formula/equation to stop here.

I'm open to criticism, correction, guidance!

Answer 3

Use the Pythagorean Theorem.