# Help with this hard geometry problem!

• Oct 19th 2009, 10:38 PM
xxravenxx
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.
• Oct 20th 2009, 05:21 AM
Soroban
Hello, xxravenxx!

Quote:

14. Let $\displaystyle ABC$ be a right triangle: .$\displaystyle \angle C = 90^o.$
A circle $\displaystyle O$ has side $\displaystyle AC$ as its diameter, meets hypotenuse $\displaystyle AB$ at $\displaystyle E.$
A tangent line to $\displaystyle O$ at $\displaystyle E$ meets side $\displaystyle BC$ at point $\displaystyle D.$

Prove that $\displaystyle \Delta BDE$ is isosceles.

Code:

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

Draw radius $\displaystyle OE.$
Since $\displaystyle OA = OE,\;\Delta AOE$ is isosceles.
. . $\displaystyle \angle OAE = \angle OEA = \theta$

Since $\displaystyle D{E}F$ is tangent at $\displaystyle E,\;\angle OEF = 90^o.$
. . Hence, $\displaystyle \angle OEA$ and $\displaystyle \angle AEF$ are complementary.
. . Let $\displaystyle \angle AEF = \theta'$

$\displaystyle \angle AEF$ and $\displaystyle \angle DEB$ are vertical angles: .$\displaystyle {\color{blue}\angle DEB = \theta'}$

In $\displaystyle \Delta ABC,\;\angle CAB= \theta$ and $\displaystyle \angle ABC$ are complementary.
. . Hence: .$\displaystyle {\color{blue}\angle ABC = \theta'}$

In $\displaystyle \Delta BDE\!:\;\angle DEB = \angle EBD = \theta'$

Therefore, $\displaystyle \Delta BDE$ is isosceles.

• Oct 20th 2009, 10:07 PM
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. (Rofl)
• Oct 21st 2009, 03:35 AM
mr fantastic
Quote:

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. (Rofl)

*Ahem* It's better than nothing ..... Be thankful (and I see you were (Clapping)) for small favours.
• Oct 27th 2009, 01:21 AM
xxravenxx
Quote:

Originally Posted by mr fantastic
*Ahem* It's better than nothing ..... Be thankful (and I see you were (Clapping)) 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.