1. ## Deductive geometry, tangents

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

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

2. Let T be the center of the circle.

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

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

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

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

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

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

Because, from there, I didn't understand.

Thanks, BG

5. Originally Posted by BG5965
$m\left( {\angle EBD} \right) = \frac{{m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)}}{2}$
$m\left( {\angle DEB} \right) = \frac{{m\left( {\widehat{EA}} \right)}}{2}\;\& \,m\left( {\widehat{EA}} \right) = m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)$
But that means that $m\left( {\angle DEB} \right) = m\left( {\angle EBD} \right)\, \Rightarrow \,\angle DEB \cong \angle EBD$