《2025年基础训练大象出版社八年级数学上册人教版》

21. (10分)如图,已知在$\triangle ABC$中,$\angle A = 60^{\circ}$,$BD和CE是\triangle ABC$的两条角平分线,$BD与CE相交于点O$.求证:$BC = CD + BE$.

答案：
如图，在BC上截取$BF = BE$，连接OF;$\because \angle A = 60^{\circ}$, BD和CE是$\triangle ABC$两条的角平分线，$\therefore \angle 1 = \angle 2 = \frac{1}{2}\angle ABC$, $\angle 3 = \angle 4 = \frac{1}{2}\angle ACB$.$\therefore \angle BOC = 180^{\circ} - (\angle 2 + \angle 4) = 180^{\circ} - \frac{1}{2}(\angle ABC + \angle ACB) = 180^{\circ} - \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} + \frac{1}{2}\angle A = 90^{\circ} + \frac{1}{2}×60^{\circ} = 120^{\circ}$,$\therefore \angle BOE = \angle COD = 60^{\circ}$.在$\triangle BOE$和$\triangle BOF$中，$\left\{\begin{array}{l} BE = BF\\ \angle 1 = \angle 2\\ BO = BO\end{array}\right.$$\therefore \triangle BOE \cong \triangle BOF(SAS)$.$\therefore \angle BOE = \angle BOF = 60^{\circ}$.$\therefore \angle COF = \angle BOC - \angle BOF = 120^{\circ} - 60^{\circ} = 60^{\circ} = \angle COD$.在$\triangle OCF$和$\triangle OCD$中，$\left\{\begin{array}{l} \angle COF = \angle COD\\ CO = CO\\ \angle 4 = \angle 3\end{array}\right.$$\therefore \triangle OCF \cong \triangle OCD(ASA)$.$\therefore CF = CD$.$\because BC = CF + BF$,$\therefore BC = CD + BE$.