# Chord parallel to AB and 2CD=AB

1. Feb 9, 2004

### himanshu121

AB is the diameter of a circle CD is a chord parallel to AB and 2CD=AB. The tangent at B meets the line AC produced at E. Prove that AE=2AB.

I'm finding no way to solve this ?

Still i thought of applying AE*CE=BE2 but that is not enough to slove the pro ???

Any other Hint to solve

2. Feb 9, 2004

### matt grime

So A and C are on the same end of the chords?

Let O be the centre of the circle

the triangle OAC is equilateral, because the triangle OCD is and everything is symmetric.. well, convince yourself somehow

so angle BAE is 60,

so AB/AE = cos60 = 1/2; AE=d. QED

Opposite ends of te chords gives you...?

3. Feb 10, 2004

### MathematicalPhysicist

i think you should bulid a trapozoid with CD and AB (with two sides equals to each other and to the small base-CD) and to see that we have THE TRAINGLE BAE as a right triangle because the angle between radius and a tangent is 90.
here are my computations that brought to the answer:
BE^2=CE*AE
AB^2+BE^2=AE^2 (PYTHAGORAS THEOREM)
AB^2+CE*AE=AE^2
AB^2=AE(AE-CE)=AE*CE
CE=CD=AB/2
AB^2/CE=AE
AB^2/(AB/2)=AE
2AB=AE

so you were half right himanshu

4. Feb 11, 2004

### MathematicalPhysicist

i have an illustration of the triangle and trapaoid the problem is my scanner doesnt work so i cant upload it here, sorry.

5. Feb 12, 2004

### himanshu121

Ya i got it
I was half way write upto formation of equation and it just required manipulations & rearrangement

Ya i convinced myself for equilateral triangle
Thnxs