2007-I-17

$\angle ABG = \angle DBG$.

In $\Delta ABG$ and $\Delta DBG$,

$\begin{array}{ll}
AB=BD & \text{(given)} \\
BG=BG & \text{(common side)} \\
\angle ABG=\angle DBG & \text{(proved)}
\end{array}$

$\therefore \Delta ABG \cong \Delta DBG$. (S.A.S.)

$\angle BAE = \angle IAG$.

Since $\Delta ABG \cong \Delta DBG$,

$\begin{array}{ll}
\angle BGA= \angle BGD & \text{(corr. sides, $\cong\Delta$s)} \\
\therefore \angle IGA=90^\circ &
\end{array}$

Since $AC$ is the diameter of the semi-circle $ABC$,

$\begin{array}{ll}
\angle ABC = 90^\circ & \text{($\angle$ in semi circle)}
\end{array}$

In $\Delta AGI$ and $\Delta ABE$,

$\begin{array}{ll}
\angle IAG=\angle EAB & \text{(proved)} \\
\angle IGA = \angle EBA = 90^\circ & \text{(proved)}
\end{array}$

$\begin{array}{rll}
\angle GAI & = 180^\circ – \angle IAG – \angle IGA & \text{($\angle$ sum of $\Delta$)} \\
& = 180^\circ – \angle EAB – \angle EBA & \text{(proved)} \\
& = \angle BAE & \text{($\angle$ sum of $\Delta$)}
\end{array}$

$\therefore \Delta AGI \sim \Delta ABE$. (A.A.A.)

Hence, we have

$\begin{array}{rcll}
\dfrac{AB}{AG} & = & \dfrac{BE}{GI} & \text{(corr. sides, $\sim\Delta$s)} \\
\dfrac{GI}{AG} & = & \dfrac{BE}{AB} &
\end{array}$

$\begin{array}{rcl}
G & = & \left( \dfrac{-25+ 11}{2}, 0 \right) \\
& = & (-7,0)
\end{array}$

$\begin{array}{rcl}
\dfrac{GI}{AG} & = & \dfrac{BE}{AB} \\
\dfrac{GI}{AG} & = & \dfrac{1}{2} \\
AG & = & 2GI \\
-7 – (-25) & = & 2y \\
y & = & 9
\end{array}$

$\therefore I=(-7,9)$.

Note that $AD$ is the tangent at $G$ of the inscribed circle of $\Delta ABD$ and $IG \perp AD$, then $IG$ is a radius of the inscribed circle of $\Delta ABD$. Hence, the equation of the inscribed circle of $\Delta ABD$ is

$\begin{array}{rcl}
(x-(-7))^2 + (y-9)^2 & = & (9-0)^2 \\
(x+7)^2 + (y-9)^2 & = & 9^2
\end{array}$