《2025年拔尖特训八年级数学上册人教版》

10. 在$\triangle ABC$中，记$\angle BAC=\alpha$，$\angle ACB=\beta$。
(1)如图①，若$AP$平分$\angle BAC$，$BP$，$CP$分别平分$\triangle ABC$的外角$\angle CBM$和$\angle BCN$，$BD\perp AP$于点$D$。
①用含$\alpha$的式子表示$\angle BPC$的度数。
②用含$\beta$的式子表示$\angle PBD$的度数。
(2)如图②，若$P$为$\triangle ABC$的三条内角平分线的交点，且$BD\perp AP$于点$D$。
①请补全图形。
②(1)中的两个结论是否发生变化？如果不变，请说明理由；如果变化，请直接写出正确的结论。

答案：
10.
(1)①$\because BP$,$CP$分别平分$\triangle ABC$的外角$\angle CBM$和$\angle BCN$,$\therefore \angle PBC = \angle PBM = \frac{1}{2}\angle CBM = \frac{1}{2}(\angle BAC + \angle ACB) = \frac{1}{2}(\alpha + \beta)$,$\angle BCP = \frac{1}{2}\angle BCN = \frac{1}{2}(180^{\circ} - \angle ACB) = \frac{1}{2}(180^{\circ} - \beta)$.$\therefore \angle BPC = 180^{\circ} - \angle PBC - \angle BCP = 180^{\circ} - \frac{1}{2}(\alpha + \beta) - \frac{1}{2}(180^{\circ} - \beta) = 90^{\circ} - \frac{1}{2}\alpha$.
②$\because BD \perp AP$,$\therefore \angle BDP = 90^{\circ}$.在$Rt\triangle PBD$中,$\angle PBD = 90^{\circ} - \angle BPD$.$\because AP$平分$\angle BAC$,$\therefore \angle BPD = \angle PBM - \angle BAP = \angle PBM - \frac{1}{2}\angle BAC = \frac{1}{2}(\alpha + \beta) - \frac{1}{2}\alpha = \frac{1}{2}\beta$.$\therefore \angle PBD = 90^{\circ} - \frac{1}{2}\beta$.
(2)①补全图形如图所示.②
(1)中的两个结论发生了变化.$\because \angle BAC = \alpha$,$\angle ACB = \beta$,$\therefore \angle ABC + \angle ACB = 180^{\circ} - \alpha$,$\angle ABC + \angle BAC = 180^{\circ} - \beta$.$\because P$为$\triangle ABC$的三条内角平分线的交点,$\therefore \angle PBC = \frac{1}{2}\angle ABC$,$\angle PCB = \frac{1}{2}\angle ACB$.$\therefore \angle PBC + \angle PCB = \frac{1}{2}(\angle ABC + \angle ACB)$.$\therefore \angle BPC = 180^{\circ} - (\angle PBC + \angle PCB) = 180^{\circ} - \frac{1}{2}(\angle ABC + \angle ACB) = 180^{\circ} - \frac{1}{2}×(180^{\circ} - \alpha) = 90^{\circ} + \frac{1}{2}\alpha$.$\because$由题意知,$BD \perp AD$,$\therefore \angle PDB = 90^{\circ}$.$\because$易知$\angle BPD = \angle ABP + \angle BAP = \frac{1}{2}(\angle ABC + \angle BAC) = \frac{1}{2}(180^{\circ} - \beta) = 90^{\circ} - \frac{1}{2}\beta$,$\therefore \angle PBD = 180^{\circ} - \angle PDB - \angle BPD = 180^{\circ} - 90^{\circ} - (90^{\circ} - \frac{1}{2}\beta) = \frac{1}{2}\beta$.