“微信扫一扫”进入题库练习及模拟考试
如图,\(\Delta ABC\)中,\(AB = AC\),\(AD\)平分\(\angle BAC\),射线\(AN\)平分外角\(\angle CAM\),过点\(C\)作\(CE \bot AN\)于点\(E\),求证:四边形\(ADCE\)是矩形。
参考答案:证明:
\(\because AB = AC\),
\(AD\)平分\(\angle BAC\),
\(\therefore AD \bot BC\),
\(\angle CAD = \frac{1}{2}\angle BAC\),
\(\therefore \angle ADC = 90^\circ \)。
\(\because AN\)是\(\Delta ABC\)外角的平分线,
\(\therefore \angle CAE \)
\(= \frac{1}{2}\angle CAM\)。
\(\because \angle BAC + \angle CAM \)
\(= 180^\circ \),
\(\therefore \angle DAN \)
\(= \angle CAD + \angle CAE\)
\( = \frac{1}{2}(\angle BAC + \angle CAM) \)
\(= 90^\circ \)。
\(\because CE \bot AN\),
\(\therefore \angle CEA = 90^\circ \),
\(\therefore \)四边形\(ADCE\)为矩形。