借您的圖
作\(\overline{MD}\bot \overline{BC}\)於D，\(\overline{ME}\bot \overline{AB}\)於E，\(\overline{NJ}\bot \overline{BC}\)於J，\(\overline{NK}\bot \overline{AC}\)於K
設\(\overline{DN}\)交\(\overline{PI}\)於T
\(\begin{align}
  & \overline{PG}=\frac{\overline{MP}}{\overline{MN}}\times \overline{NK}=\frac{\overline{MP}}{\overline{MN}}\times \overline{NJ}=\frac{\overline{DI}}{\overline{DJ}}\times \overline{NJ}=\overline{IT} \\
& \overline{PH}=\frac{\overline{NP}}{\overline{MN}}\times \overline{ME}=\frac{\overline{NP}}{\overline{MN}}\times \overline{MD}=\overline{PT} \\
& \overline{PI}=\overline{PT}+\overline{IT}=\overline{PG}+\overline{PH} \\
\end{align}\)

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 ]

上一題結論\(\overline{PI}=\overline{PG}+\overline{PH}\)
作\(\overline{EX}\bot \overline{AB}\)於X，\(\overline{EY}\bot \overline{BC}\)於Y，\(\overline{EZ}\bot \)於Z
取\(\overline{PN}=\overline{EN}\)，則\(\overline{PH}=\overline{EX}\)
\(\begin{align}
  & \overline{PI}+\overline{EY}=2\overline{NJ}=2\overline{NK}=\overline{PG}+\overline{EZ} \\
& \overline{PG}+\overline{PH}+\overline{EY}=\overline{PG}+\overline{EZ} \\
& \overline{EX}+\overline{EY}=\overline{EZ} \\
\end{align}\)