题目
Geogebra作图
证明：
作出完全四边形ABECFD的密克尔点G，连接OG，作EF中点M，过点O作LA垂线并交于点N；
设(angle OLN=alpha,angle FLP=eta,angle ALF=gamma)
(先证明G在EF上且OGot EF)
[angle EGC+angle EGF=angle ABC+angle ADC=180^circ
]
[OE^2-OF^2=EBcdot EA-FDcdot FA=EGcdot EF-FGcdot FE=EG^2-FG^2
]
设圆(O)半径为(r)
[OG^2=OE^2-EG^2=EGcdot EF+r^2-EG^2=EGcdot FG+r^2
]
计算点(L)对圆(O)的幂
[egin{align}
&LAcdot(2LN-LA)=OL^2-r^2=OG^2+GL^2-r^2\
Leftrightarrow &2LAcdot LN-LA^2=OG^2+LG^2-r^2
\Leftrightarrow &2LAcdot LN=LA^2+EGcdot FG+r^2+LG^2-r^2=LMcdot LG
end{align}]
即点ANGM四点共圆
[egin{align}
0&=LNcdot LA-LGcdot LM\
&=OL cosalphacdot PL cos(gamma-eta)-OLcos(gamma-alpha)cdot PLcoseta\
&=OLcdot PL (cosalphacos(gamma-eta)-cosetacos(gamma-alpha))\
&=OLcdot PL singamma sin(eta-alpha)
end{align}]
即证(alpha=eta)