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 $.

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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?

  • 1
    $\begingroup$ CD=(3/4)AB? Is that a typo or am I misreading? You said at the beginning that CD=(1/2)AB. $\endgroup$
    – Tyler
    Jul 13, 2013 at 5:40
  • $\begingroup$ That's a typo! Sorry. Editing it. $\endgroup$ Jul 13, 2013 at 5:41
  • $\begingroup$ @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.. $\endgroup$
    – user86123
    Jul 13, 2013 at 6:24
  • $\begingroup$ 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. $\endgroup$ Jul 13, 2013 at 6:35

2 Answers 2


Name the center of the circle O. Then OCA is an equilateral triangle (OA and OC are both radii so they must be the same length) which means the angle CAB is 60 degrees. The triangle EAB is thus a "30-60-90" triangle which means its sides have a ratio of 1 to 2 to sqrt(3) with AE being the "2" side and AB being the "1" side.

  • $\begingroup$ Thanks! Marking the center of the circle was a great idea indeed :D $\endgroup$ Jul 13, 2013 at 5:53
  • $\begingroup$ I suspect that any proof will involve the center of the circle but I'm not quite sure how to prove that. $\endgroup$
    – Tyler
    Jul 13, 2013 at 5:55
  • $\begingroup$ Ok! Will keep that in mind $\endgroup$ Jul 13, 2013 at 5:57

You could extend the line BD to meet AE at F. Then prove something special about F. We know that triangle FAB is isoceles. Does that ring a bell? Going along these lines will give you a relation between AE and FB which will give you a relation between AE and BD. Furthermore, you find two similar triangles involving AE, BD and AB. Manipulating the ratios will give you the answer you need.


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