# A proof in circles.

I need help proving this problem:

$AB$ is a 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 $AE = 2AB$.

What I've got so far is this:

on extending the line $CD$ to the tangent at $B$ such that $CD$ and the tangent meet at some point $H$, I know that $CH = \dfrac 34 AB$. So from this I know that $CE = \dfrac 3 4 AE.$

How to go further?

• CD=(3/4)AB? Is that a typo or am I misreading? You said at the beginning that CD=(1/2)AB. Jul 13, 2013 at 5:40
• That's a typo! Sorry. Editing it. Jul 13, 2013 at 5:41
• @MatrixFrog.. The assumption of an equilateral triangle is false.. Since it is not clear by your deductions that AC = OC = OA.. Thus far you have only shown that at best triangle OCA is a scalene triangle.. And thus the assumption that the angle CAB is 60 degrees is false.. And the proof is not truly complete..
– user86123
Jul 13, 2013 at 6:24
• This should have been a comment. Btw although, I don't know how I forgot to see that, what you are asking for can be easily proved. Angle COD = 60deg. And triangles OAC and ODB are congruent (can be easily seen by the symmetry in the problem). Since angle AOC + COD + DOB = 180deg, and also angles AOC = DOB, we know each is 60 deg. Jul 13, 2013 at 6:35