回復 1# bch0722b 的帖子
我就不附圖了
先證MN垂直AE
AMDN四點共圓， \(\displaystyle \angle{AMN}=\angle{ADN} \)
所以 \(\displaystyle \angle{AMN}+ \angle{MAE}= \angle{ADN}+ \angle{DAN}=90^o \)
於是四邊形AMON面積= \(\displaystyle \frac{1}{2} AO \times MN \)
又圓AMDN的直徑為AD
所以 \(\displaystyle MN=AD \sin\angle{MAN} \)
\(\displaystyle \Delta ABO ~ \Delta ADC \)
\(\displaystyle AO \times AD=AB \times AC \)
四邊形AMON面積= \(\displaystyle \frac{1}{2} AO \times MN \)
\(\displaystyle =\frac{1}{2} AO \times AD \sin\angle{MAN} \)
\(\displaystyle =\frac{1}{2} AB \times AC \sin\angle{MAN} \)
[ 本帖最後由 lyingheart 於 2014-6-8 03:04 PM 編輯 ]
