回復 1# deca0206 的帖子
再借一次您的第(二)個圖
作\overline{EX}\bot \overline{AB}於X,\overline{EY}\bot \overline{BC}於Y,\overline{EZ}\bot \overline{CA}於Z
由上一題可證出\overline{EX}+\overline{EY}=\overline{EZ}
\begin{align}
& 2\Delta AEB=\overline{AB}\times \overline{EX}=\overline{EA}\times \overline{EB}\times \sin AEB=\overline{EA}\times \overline{EB}\times \sin ACB \\
& \overline{EX}=\overline{EA}\times \overline{EB}\times \frac{\sin ACB}{\overline{AB}} \\
\end{align}
同理
\begin{align}
& \overline{EY}=\overline{EB}\times \overline{EC}\times \frac{\sin BAC}{\overline{BC}} \\
& \overline{EZ}=\overline{EC}\times \overline{EA}\times \frac{\sin CBA}{\overline{CA}} \\
& \overline{EX}+\overline{EY}=\overline{EZ} \\
& \overline{EA}\times \overline{EB}\times \frac{\sin ACB}{\overline{AB}}+\overline{EB}\times \overline{EC}\times \frac{\sin BAC}{\overline{BC}}=\overline{EC}\times \overline{EA}\times \frac{\sin CBA}{\overline{CA}} \\
& \overline{EA}\times \overline{EB}+\overline{EB}\times \overline{EC}=\overline{EC}\times \overline{EA} \\
& \frac{1}{\overline{EC}}+\frac{1}{\overline{EA}}=\frac{1}{\overline{EB}} \\
\end{align}
這題實在很漂亮,不太適合小弟這種有點年紀的中年大叔啊
[ 本帖最後由 thepiano 於 2015-10-2 09:22 PM 編輯 ]