問題:
題解:
Joint $BT$ and $DT$.
Let $MT=a$, $ME=b$, $MD=r$ and $MB=s$
Our aim is to prove $a=b$.
Consider $\triangle BMT$ and $\triangle TMD$,
$\angle BMT= \angle TMB$ (commond $\angle$)
$\angle TBM= \angle DTM$ ($\angle$ in alt. segment)
Thus, $\triangle BMT \sim \triangle TMD$ (AA)
By corr. sides $\sim \triangle s$, we have,
$\begin{align*}
\frac{MT}{MD}&=\frac{MB}{MT}\\
\frac{a}{r}&=\frac{s}{a}\\
a&=\sqrt{rs}
\end{align*}$
Consider $\triangle BME$ and $\triangle EMD$,
$\angle BME= \angle EMD$ (commond $\angle$)
$\angle MBE= \angle A$ ($\angle s$ in the same segment)
$\angle A= \angle MED$ (alt. $\angle s$, $EM//AC$)
so, $\angle MBE= \angle MED$
Thus, $\triangle BME \sim \triangle EMD$ (AA)
By corr. sides $\sim \triangle s$, we have,
$\begin{align*}
\frac{ME}{MD}&=\frac{MB}{ME}\\
\frac{b}{r}&=\frac{s}{b}\\
b&=\sqrt{rs}=a
\end{align*}$
Therefore, $MT=ME$.
$Q.E.D.$
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