Thread: Help with this hard geometry problem!

Problem 14
Let ABC be a right angled triangle. A circle Г have side AC as its diameter meets hypotenuse AB at point E. A tangent line to Г at point E meets side BC at point D. Prove that a triangle BDE is isosceles.

Hello, xxravenxx!

14. Let be a right triangle: .
A circle has side as its diameter, meets hypotenuse at
A tangent line to at meets side at point

Prove that is isosceles.

Code:

          F
           o
            \
           θ'\
      A o *   \
        | θ * *\ E
        |     θ o
        |     *  *  *
        |   *     \ θ ' *
        | *       *\        *
      O *         * \           *
        |         *  \              *
        |             \                 *
        |        *     \                    *
        |       *       \                       *
        |     *          \                       θ' *
      C o * - - - - - - - o - - - - - - - - - - - - - - o B
                            D

Draw radius
Since is isosceles.
. .

Since is tangent at
. . Hence, and are complementary.
. . Let

and are vertical angles: .


In and are complementary.
. . Hence: .


In

Therefore, is isosceles.

Awesome. Thanks heaps. But a little advice - try to draw a diagram through paint or something instead of typing it up. It makes it a little hard to understand. But I got there.

Thank you.

Originally Posted by xxravenxx

Awesome. Thanks heaps. But a little advice - try to draw a diagram through paint or something instead of typing it up. It makes it a little hard to understand. But I got there.

Thank you.

*Ahem* It's better than nothing ..... Be thankful (and I see you were ) for small favours.

Originally Posted by mr fantastic

*Ahem* It's better than nothing ..... Be thankful (and I see you were ) for small favours.

No no, I'm just saying. It would be easier for you to draw up a pic in paint then type up a picture.