平面幾何基本功：用導角法解決若干問題

引理1 如圖, 設銳角\(\small \triangle ABC\)的外接圓為\(\small\Omega, X,Y,Z\)分別是劣弧\(\small\mathop{BC}\limits^\frown,\mathop{AC}\limits^\frown,\mathop{AB}\limits^\frown\)的中點.證明：\(\small\triangle XYZ\)的垂心是\(\small\triangle ABC\)的內心.

分析：易知\(\small AX,BY,CZ\)是角平分線, 所以只需證明\(\small \triangle ABC\)的內心\(\small I\)就是\(\small \triangle XYZ\)的垂心. 換言之, 即證明\(\small AX\perp YZ,BY\perp XZ,CZ\perp XY\), 於是可以將垂直關係轉化為兩角和等於90\(^\circ\), 接下來利用導角法完成即可.

證：因為\(\small \angle XZY=\angle XBY=\angle XBC+\angle CBY=\angle XAC+\angle CBY\),所以我們有,

\[\small \begin{align*} \angle XZY+\angle AXZ&=\angle XAC+\angle CBY+\angle AXZ\\ &=\angle XAC+\angle CBY+\angle ACZ\\ &=\frac{1}{2}(\angle BAC+\angle CBA+\angle ACB)\\ &=90^\circ, \end{align*} \]

這說明\(\small AX\perp YZ\),同理可得\(\small BY\perp XZ,CZ\perp XY\).所以, \(\small \triangle ABC\)的內心\(I\)就是\(\small \triangle XYZ\)的垂心.

問題1(JMO2011/5) 點\(\small A,B,C,D,E\)在圓\(\small \omega\)上, 而點\(\small P\)在圓\(\small \omega\)外.這些點滿足：(1)直線\(\small PB,PD\)與圓\(\small \omega\)相切;(2)\(\small P,A,C\)共線;(3)\(\small \overline{DE}\parallel \overline{AC}\).證明：\(\small \overline{BE}\) 平分\(\small \overline{AC}\).

分析：\(\small OF\perp AC\)等價於\(\small F\)為\(\small AC\)中點, 將線段數量關係轉化為位置關係是解題關鍵.

證：
根據題目條件(3)知\(\small \angle BFP=\angle BED\),又因為\(\small \angle BED=\frac{1}{2}\angle BOD=\angle BOP\),所以\(\small \angle BFP=\angle BOP\),這說明\(\small B,O,F,P\)四點共圓.於是我們有,

\[\small\angle FOP+\angle OPF=\angle FBP+\angle OBF=90^\circ, \]

即\(\small OF\perp AC\),所以\(\small F\)為\(\small AC\)中點.

引理2 如圖,設銳角\(\small \triangle ABC\)中, \(\small \overline{BE},\overline{CF}\)是高, \(\small M\)是\(\small \overline{BC}\)的中點.證明:\(\small \overline{ME},\overline{MF}\)和過\(\small A\)與\(\small \overline{BC}\)平行的直線均與圓\(\small (AEF)\)相切.

證：設\(\small \overline{BE}\)與\(\small \overline{CF}\)交於點\(\small D\).因為\(\small \angle DFA=\angle DEA=90^\circ\), 所以\(\small A,F,D,E\)四點共圓, 且\(\small \overline{AD}\)為直徑, 由題意知\(\small D\)為\(\small \triangle ABC\)垂心, 這說明\(\small \overline{AD}\perp \overline{BC}\), 所以過\(\small A\)與\(\small \overline{BC}\)平行的直線垂直於\(\small \overline{AD}\), 所以該直線與圓\(\small (AEF)\)相切. 接下來證明\(\small \overline{MF}\)與\(\small \overline{OF}\)垂直.

首先注意到\(\small E,F,B,C\)四點共圓且\(\small M\)是\(\small \overline{BC}\)中點, 那麼\(\small \overline{MF}=\overline{MC}\), \(\small \angle BCF=\angle BEF=\angle DAF\). 所以,

\[\small \angle MFO=\angle MFC+\angle DFO=\angle DAF+\angle ODF=90^\circ. \]

這說明\(\small \overline{MF}\)與\(\small \overline{OF}\)垂直, 即\(\small \overline{MF}\)與圓\(\small (AEF)\)相切. 同理可得, \(\small \overline{ME}\)與圓\(\small (AEF)\)相切.

引理3(內切圓弦上的直角) 如圖, 設\(\small \triangle ABC\)的內切圓在邊\(\small \overline{BC},\overline{AC},\overline{AB}\)上的切點分別是\(\small D,E,F\), 內切圓圓心為\(\small I\). 設\(\small M,N\)分別是$\overline{BC},\overline{AC} \(的中點.射線\)\small BI\(與直線\)\small EF\(相交於\)\small K\(. 證明：\)\small \overline{BK}\perp \overline{CK}\(, 並且\)\small K\(在直線\)\small MN$上.

分析：不妨將結論換一種方向, 假設射線\(\small BI\)與直線\(\small MN\)相交於\(\small K'\), 然後證明\(\small K'\)在直線\(\small EF\)上.

證： 假設射線\(\small BI\)與直線\(\small MN\)相交於\(\small K'\). 因為\(\small MN\parallel AB\),所以\(\small \angle BK'M=\angle ABK'=\angle K'BM\), 這說明\(\small \triangle BMK'\)是等腰三角形, 即有\(\small K'M=BM\), 又因為\(\small BM=CM\), 所以\(\small \angle BK'C=90^\circ\), 即\(\small \overline{BK}\perp \overline{CK}\). 下面證明\(\small K'\)在直線\(\small EF\)上.

\[\small \begin{align*} \overline{EN}&=\frac{1}{2}\overline{AC}-\overline{AE}\\ \overline{NK}&=\overline{MK}-\overline{MN}=\frac{1}{2}\overline{BC}-\overline{AB}\\ \overline{EN}-\overline{NK}&=\frac{1}{2}\overline{AC}-\overline{AE}+\frac{1}{2}(\overline{AB}-\overline{BC})\\ &=\frac{1}{2}\overline{AC}-\overline{AE}+\frac{1}{2}(\overline{AE}-\overline{CE})\\ &=\frac{1}{2}\overline{AC}-\frac{1}{2}\overline{AE}-\frac{1}{2}\overline{CE}\\ &=\frac{1}{2}\overline{AC}-\frac{1}{2}\overline{AC}=0^\circ, \end{align*} \]

於是\(\small \overline{EN}=\overline{NK}\), 又因為\(\small \angle ENK'=\angle MNC=\angle A\), 所以\(\small \triangle NEK'\sim \triangle AEF\),可得\(\small \angle NEK'=\angle AEF\), 注意到\(\small A,E,N\)共線, 故\(\small E,F,K'\)共線, 這說明射線\(\small BI\)與直線\(\small EF\)交於點\(\small K'\), 所以\(\small K'\)與\(\small K\)為同一點.

問題2(加拿大1997/4) 點\(\small O\)在平行四邊形\(\small ABCD\)內部, 使得\(\small \angle AOB+\angle COD=180^\circ\).證明：\(\small \angle OBC=\angle ODC\).

證： 將\(\small \triangle ABO\)平移至\(\small \triangle DCE\),\(\small \angle COD+\angle DEC=\angle COD+\angle AOB=180^\circ\),所以\(\small O,D,E,C\)四點共圓, 易得\(\small \angle ODC=\angle OEC\).因為\(\small \triangle BOC\simeq \triangle ECO\), 所以\(\small \angle OBC=\angle OEC=\angle ODC\).

問題3(IMO2006/1) 設\(\small \triangle ABC\)的內心為\(\small I,P\)在三角形內部, 滿足
$$\small \angle PBA+ \angle PCA=\angle PBC+\angle PCB.$$
證明：\(\small AP\ge AI\),等號成立當且僅當\(\small P=I\).

分析：對等式兩邊移項, \(\small \angle PBA-\angle PBC=\angle PCB-\angle PCA\),由等式兩邊的符號相同可推出點\(\small P\)落在\(\small \triangle IBC\)外, 合理猜測\(\small P,B,C,I\)四點共圓, 結合雞爪定理可驗證滿足題意.

證：

\[\small \begin{align*} \angle PBA+\angle PCA&=\angle IBA-\angle IBP+\angle ICA+\angle ICP\\ &=\angle IBC-\angle IBP+\angle ICB+\angle ICP\\ &=\angle IBC+\angle ICB-\angle IBP+\angle ICP\\ &=180^\circ -\angle BIC-\angle IBP+\angle ICP\\ \angle PBC+\angle PCB&=180^\circ -\angle BPC\\ \Rightarrow \angle BPC+\angle ICP&=\angle BIC+\angle IBP \\ \Rightarrow 180^\circ-\angle PDB-\angle IBP&+\angle ICP=180^\circ-\angle IDC-\angle ICP+\angle IBP, \end{align*} \]

因為\(\small \angle PDB=\angle IDC\),所以上式兩邊抵消可得\(\small \angle IBP= \angle ICP\), 這說明\(\small P,B,C,I\)四點共圓. 由雞爪定理可知, \(\small AP\ge AI\), 等號成立當且僅當\(\small P=I\).

注： 雞爪定理在證明過程中不能直接使用, 需證明後才能使用, 因(比較懶)篇幅原因證明略.

引理4(西姆松線) 如圖, 設\(\small P\)是\(\small \triangle ABC\)外接圓上任一點, \(\small X,Y,Z\)分別是從\(\small P\)到直線\(\small BC,CA,AB\)的投影. 證明：\(\small X,Y,Z\)共線.

證： 根據垂直關係得到點\(\small P\)在三個圓\(\small (YZA),(ZXB),(XYC)\)上, 易得

\[\small \begin{align*} \measuredangle PYZ & =\measuredangle PAZ=\measuredangle PAB\\ &=\measuredangle PCB=\measuredangle PCX=\measuredangle PYX. \end{align*} \]

所以\(\small X,Y,Z\)三點共線.

問題4(USAMO 2010/1) 設凸五邊形\(\small AXYZB\)內接於以\(\small AB\)為直徑的半圓. 記\(\small P,Q,R,S\)分別是\(\small Y\)到直線\(\small AX,BX,AZ,BZ\)上的投影. 證明：直線\(\small PQ\)和\(\small RS\)形成的銳角是\(\small \angle XOZ\)的一半, 其中\(\small O\)是線段\(\small AB\)的中點.

證： 作\(\small YC\perp AB\)於\(\small C\), 根據西姆松引理可得\(\small P,Q,C\)與\(\small S,R,C\)三點共線. 由題目條件易知\(\small A,P,Y,C\)四點共圓, \(\small C,Y,S,B\)四點共圓, 於是我們有\(\small \angle PCY=\angle XAY=\angle XBY,\angle SCY=\angle ZBY\), 此時,

\[\small \begin{align*} \small \angle PCS&=\angle PCY+\angle SCY\\ &=\angle XBY+\angle ZBY\\ &=\angle ZBX=\frac{1}{2}\angle XOZ. \end{align*} \]

問題5(IMO 2013/4) 設銳角\(\small \triangle ABC\)的垂心為\(\small H,W\)是邊\(\small \overline{BC}\)上一點, 位於\(\small B,C\)中間. 點\(\small M,N\)分別是從\(\small B,C\)引出的三角形的高的垂足. \(\small \omega_1\)是\(\small \triangle BWN\)的外接圓, 點\(\small X\)滿足\(\small \overline{WX}\)是\(\small \omega_1\)的直徑. 類似地, 點\(\small Y\)滿足\(\small \overline{MY}\)是\(\small \triangle CWM\)外接圓\(\small \omega_2\)的直徑. 證明：點\(\small X,Y,H\)共線.

證： 設\(\small \omega_1\)與\(\small \omega_2\)相交於另一點\(\small D\), 因為\(\small \angle XDM=\angle YDW=90^\circ\), 所以\(\small X,D,Y\)三點共線. 下面只需證明\(\small H\)在直線\(\small XD\)上. 注意到\(\small \angle ADN+\angle ADM=\angle ABC+\angle ACB=180^\circ-\angle BAC\), 所以\(\small A,N,D,M\)四點共圓, 又因為\(\small A,N,H,M\)四點共圓, 所以\(\small A,N,H,D,M\)五點共圓, 此時\(\small AH\)是直徑, 所以\(\small AD\perp HD\),所以\(\small H\)在直線\(\small XD\)上. 於是\(\small X,H,Y\)三點共線.

注：\(\small D\)稱作密克點，後期文章補充(給自己挖坑).

問題6(IMO 1985/1]) 某個圓的圓心在圓內接四邊形\(\small ABCD\)的邊\(\small \overline{AB}\)上, 並且與另外三邊相切. 證明：\(\small AD+BC=AB\).

證： 在\(\small AB\)上作點\(\small T\)滿足\(\small AD=AT\), 那麼\(\small \angle DTA=\frac{1}{2}(\pi -\angle DAB)=\frac{1}{2}\angle DCB=\angle DCE\),這說明\(\small D,C,E,F\)四點共圓, 於是我們有\(\small \angle CTB=\angle CDE=\frac{1}{2}\angle CDA=\frac{1}{2}(\pi-\angle ABC)\),這說明\(\small TB=BC\), 所以\(\small AD+BC=AT+TB=AB\).