答疑:一道平面几何题(2)

同事代她朋友的儿子问我两道平面几何题,第一题我已经给出解答,下面我们来求解第2个平面几何题:

题目:已知$\triangle ABC$中,$H$是$\triangle ABC$的垂心,直线$BH$交直线$AC$于点$D$.$M$为$AH$的中点.$O$是$\triangle ABH$的外心.直线$CM,DO$交于点$G$.连接$HG$,过点$C$作$HG$的平行线交$AH$延长线于点$E$.连接$DE$.求证:$DO\perp DE$.

此图像的alt属性为空;文件名为20190407-1.png

此图像的alt属性为空;文件名为20190407-2.png

证明:如图2,设$\triangle ABH$中,$\angle ABH=\alpha$,$\angle BAH=\beta$,设$\triangle ABH$的外接圆半径为$R$.设直线$OD$与直线$BC$交于点$F$,直线$OD$与直线$AB$交于点$P$.设线段$BH$的中点为$Q$.过点$E$作直线$AC$的垂线,设垂足为$K$.先用$\alpha,\beta,R$表示出若干线段的长度和角的度数.易得
$$
AH=2R\sin\alpha,BH=2R\sin\beta,
$$
$$
\angle OBA=\angle OAB=\angle HAD=\angle HBC=90^{\circ}-(\alpha+\beta).
$$

\begin{align*}
\tan\angle BDP&=\frac{OQ}{QD}=\frac{OQ}{QH+HD}\\&=\frac{R\sin\angle
OBH}{\frac{1}{2}BH+AH\sin\angle
HAD}\\&=\frac{R\sin(90^{\circ}-\beta)}{R\sin\beta+2R\sin\alpha\sin[90^{\circ}-(\alpha+\beta)]}\\&=\frac{\cos\beta}{\sin\beta+2\sin\alpha\cos(\alpha+\beta)}
\\&=\frac{\cos\beta}{\sin(2\alpha+\beta)}.
\end{align*}

\begin{align*}
\frac{AP}{PB}&=\frac{S_{\triangle DAP}}{S_{\triangle
DBP}}=\frac{\frac{1}{2}DA\cdot DP\sin\angle
ADP}{\frac{1}{2}DB\cdot DP\sin\angle BDP}\\&=\frac{\frac{1}{2}DA\cdot DP\cos\angle
BDP}{\frac{1}{2}DB\cdot DP\sin\angle BDP}\\&=\frac{DA}{DB\tan\angle
BDP}
\\&=\frac{\tan\alpha}{\tan\angle BDP}
\\&=\frac{\tan\alpha\sin(2\alpha+\beta)}{\cos\beta}.
\end{align*}

$$
\frac{CD}{DA}=\frac{CD/BD}{DA/BD}=\frac{\tan \angle DBC}{\tan \angle
ABD}=\frac{1}{\tan\alpha\tan(\alpha+\beta)}.
$$由梅涅劳斯定理,
$$
\frac{AP}{PB}\cdot \frac{BF}{FC}\cdot \frac{CD}{DA}=1,
$$

$$
\frac{FC}{BF}=\frac{AP}{PB}\cdot \frac{CD}{DA}=\frac{\sin(2\alpha+\beta)}{\cos\beta\tan(\alpha+\beta)}
$$
因为
$$
BD=BH+HD=2R\sin\beta+2R\sin\alpha\cos(\alpha+\beta)=2R\sin(\alpha+\beta)\cos\alpha,
$$
所以
$$
BC=\frac{BD}{\cos\angle DBC}=\frac{2R\sin(\alpha+\beta)\cos\alpha}{\sin(\alpha+\beta)}=2R\cos\alpha.
$$
于是
$$
\frac{FC}{BF}=\frac{FC}{BC+CF}=\frac{FC}{2R\cos\alpha+FC}=\frac{\sin(2\alpha+\beta)}{\cos\beta\tan(\alpha+\beta)},
$$
解得
\begin{align*}
FC&=\cfrac{2R\cos\alpha}{\cfrac{\cos\beta\tan(\alpha+\beta)}{\sin(2\alpha+\beta)}-1}\\&=
\frac{2R\cos\alpha\sin(2\alpha+\beta)}{\cos\beta\tan(\alpha+\beta)-\sin(2\alpha+\beta)}
\\&=\frac{2R\cos\alpha\sin(2\alpha+\beta)\cos(\alpha+\beta)}{\cos\beta\sin(\alpha+\beta)-\sin(2\alpha+\beta)\cos(\alpha+\beta)}
\\&=\frac{2R\cos\alpha\sin(2\alpha+\beta)\cos(\alpha+\beta)}{\cfrac{\sin(\alpha+2\beta)+\sin\alpha}{2}-\cfrac{\sin(3\alpha+2\beta)+\sin\alpha}{2}}
\\&=\frac{2R\cos\alpha\sin(2\alpha+\beta)\cos(\alpha+\beta)}{\cfrac{\sin(\alpha+2\beta)-\sin(3\alpha+2\beta)}{2}}
\\&=\frac{2R\cos\alpha\sin(2\alpha+\beta)\cos(\alpha+\beta)}{-\sin\alpha\cos(2\alpha+2\beta)}.
\end{align*}
由于$OM=R\cos\alpha$,故
$$
\frac{MH}{HE}=\frac{MG}{GC}=\frac{OM}{FC}=-\frac{\sin\alpha\cos(2\alpha+2\beta)}{2\sin(2\alpha+\beta)\cos(\alpha+\beta)}.
$$
由于$MH=R\sin\alpha$,解得
$$
HE=-\frac{2R\sin(2\alpha+\beta)\cos(\alpha+\beta)}{\cos(2\alpha+2\beta)},
$$
下面先用$HE$表示出$\tan\angle EDK$.
$$
EK=AE\sin\angle EAK=(AH+HE)\sin\angle EAK=(2R\sin\alpha+HE)\cos(\alpha+\beta),
$$
$$
AK=AE\cos\angle EAK=(AH+HE)\cos\angle EAK=(2R\sin\alpha+HE)\sin(\alpha+\beta).
$$
所以
$$
DK=AK-AD=(2R\sin\alpha+HE)\sin(\alpha+\beta)-2R\sin\alpha\sin(\alpha+\beta)=HE\sin(\alpha+\beta),
$$
所以
\begin{align*}
\tan\angle
EDK=\frac{EK}{DK}&=\frac{(2R\sin\alpha+HE)\cos(\alpha+\beta)}{HE\sin(\alpha+\beta)}
\\&=\frac{2R\sin\alpha}{HE\tan(\alpha+\beta)}+\frac{1}{\tan(\alpha+\beta)}
\\&=-\frac{\sin\alpha\cos(2\alpha+2\beta)}{\sin(2\alpha+\beta)\sin(\alpha+\beta)}+\frac{1}{\tan(\alpha+\beta)}
\\&=-\frac{\sin\alpha\cos(2\alpha+2\beta)}{\sin(2\alpha+\beta)\sin(\alpha+\beta)}+\frac{\cos(\alpha+\beta)\sin(2\alpha+\beta)}{\sin(\alpha+\beta)\sin(2\alpha+\beta)}
\\&=\frac{\sin(2\alpha+\beta)\cos(\alpha+\beta)-\sin\alpha\cos(2\alpha+2\beta)}{\sin(2\alpha+\beta)\sin(\alpha+\beta)}
\\&=\frac{[\sin(3\alpha+2\beta)+\sin\alpha]-[\sin(3\alpha+2\beta)-\sin(\alpha+2\beta)]}{2\sin(2\alpha+\beta)\sin(\alpha+\beta)}
\\&=\frac{\sin\alpha+\sin(\alpha+2\beta)}{2\sin(2\alpha+\beta)\sin(\alpha+\beta)}
\\&=\frac{\sin(\alpha+\beta)\cos\beta}{\sin(2\alpha+\beta)\sin(\alpha+\beta)}
\\&=\frac{\cos\beta}{\sin(2\alpha+\beta)}
\\&=\tan\angle BDP,
\end{align*}
故$\angle BDP=\angle EDK$.由于$\angle EDK+\angle BDE=90^{\circ}$,故$\angle BDP+\angle BDE=90^{\circ}$,即$DO\perp DE$.$\Box$

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