某道數學題

其實這題用四點共圓一下就秒了，但是馮老師要我弄個不用四點共圓的方法，於是就有了下文

---

$首先看一下圖$

$答案中要用四點共圓的貌似就是得到 \angle ADH=\angle ACG$
$那不用四點共圓證到這個就好了$
$延長DA，CE交於點P,如下圖$

$要證的就是\angle 2=\angle3$
$顯而易見\angle 4=\angle PDC,\angle 5=\angle PCD (外角的內對角相等)$
$\therefore \Delta PAG \backsim \Delta PCD$
$\therefore \frac{PA}{PC}=\frac{PG}{PD}$
$\therefore \frac{PA}{PG}=\frac{PC}{PD}$
$\therefore \Delta PAC \backsim \Delta PGD$
$\therefore \angle 3= \angle 2$

方法二

$延長EC，使得CP=AG,連接DP$

$\because \angle DCP = 180^\circ - \angle ACD - \angle 1 = 120^\circ - \angle 1, \angle DAG = 120^\circ - \angle 2$
$\therefore \angle DCP = \angle DAG$
$又 \because AD=DC,AG=CP$
$\therefore \Delta DAG ≌ \Delta DCP$
$\therefore DG=DP$
$\therefore \angle 5 = \angle 4=\angle 3= 60^\circ$
然後就簡單了哈哈