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一.代数证明

$\because 对与△ACB中\angle c外接三角形是\angle BCD$
$\therefore \angle BCD=\pi -\angle C$
$\because \angle A+\angle B+\angle C=\pi$
$\therefore \angle BCD=\angle A+\angle B$
$\therefore \angle BCD>\angle A\wedge \angle BCD>\angle B$

二.几何证明

在△ABC做边BC的中点F,连接AF做角平分线,找到A于F的对称点$A^{\prime }$,连接$A^{\prime }C$
$\because AF=A^{\prime }F,BF=CF,\angle BFA=\angle CF{A}^{\prime }$
$\therefore △BFA≌△A^{\prime }FC$
$\therefore \angle B=\angle FC{A}^{\prime }$
$\because \angle FC{A}^{\prime }在\angle FCD内$
$所以\angle FCD>\angle B$
$\because \angle A+\angle B=\pi -\angle BCA=\angle DCF,\wedge \angle A+\angle B<\frac{\pi }{2}$
$\therefore \angle A<\frac{\pi }{2}-\angle B$
$\therefore \angle A<\frac{\pi }{2}-\pi +\angle BCA$
$\therefore \angle A+\frac{\pi }{2}<\angle BCA$
$\therefore \angle BCA>\angle A$