# Thread: Deductive geometry, tangents

1. ## Deductive geometry, tangents

Hi, I have a problem which I am having trouble solving.

Let $\displaystyle ABC$ be a right-angled triangle. A circle $\displaystyle T$ having side $\displaystyle AC$ as its diameter meets hypotenuse $\displaystyle AB$ at point $\displaystyle E$. A tangent line to $\displaystyle T$ at point $\displaystyle E$ meets side $\displaystyle BC$ at point $\displaystyle D$. Prove that triangle $\displaystyle BDE$ is isosceles.

2. Let T be the center of the circle.

Quadrilater CTED is cyclic. Then $\displaystyle \widehat{TCE}=\widehat{TDE}$ (1)

$\displaystyle CE\perp AB\Rightarrow\widehat{TCE}=\widehat{DBE}$ (2)

$\displaystyle \Delta TCD\equiv\Delta TED\Rightarrow\widehat{TDE}=\widehat{TDC}$ (3)

From (1), (2), (3) $\displaystyle \widehat{TDC}=\widehat{DBE}\Rightarrow TD\parallel AB\Rightarrow\widehat{TDE}=\widehat{DEB}\Rightarro w\widehat{DEB}=\widehat{DBE}$

3. I would show the $\displaystyle \angle DEB \cong \angle BED$.
This can be done by noting that both measure $\displaystyle \frac{m(\widehat{AE})}{2}$.

4. Why is $\displaystyle CE\perp AB\$?

Because, from there, I didn't understand.

Thanks, BG

5. Originally Posted by BG5965
red_dog, can you please reply, or could someone please explain my question in the above post?
$\displaystyle m\left( {\angle EBD} \right) = \frac{{m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)}}{2}$
$\displaystyle m\left( {\angle DEB} \right) = \frac{{m\left( {\widehat{EA}} \right)}}{2}\;\& \,m\left( {\widehat{EA}} \right) = m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)$
But that means that $\displaystyle m\left( {\angle DEB} \right) = m\left( {\angle EBD} \right)\, \Rightarrow \,\angle DEB \cong \angle EBD$