《2025年拔尖特训八年级数学上册人教版》

(1)若$\angle 1=20^{\circ}$，$\angle 2=30^{\circ}$，$\angle BEC=100^{\circ}$，则$\angle BDC$的度数为$150^{\circ}$。
(2)若$BE$平分$\angle ABD$，$CE$平分$\angle ACD$，$BE$与$CE$相交于点$E$，请写出$\angle BDC$，$\angle BEC$和$\angle BAC$之间的数量关系，并说明理由。
$\angle BDC + \angle BAC = 2\angle BEC$
理由:由题意,得$\angle BDC = \angle BEC + \angle 1 + \angle 2$①,$\angle BEC = \angle BAC + \angle ABE + \angle ACE$②.$\because BE$平分$\angle ABD$,$CE$平分$\angle ACD$,$\therefore \angle ABE = \angle 1$,$\angle ACE = \angle 2$.①$ - $②,得$\angle BDC - \angle BEC = \angle BEC - \angle BAC$.$\therefore \angle BDC + \angle BAC = 2\angle BEC$.
(3)如图②，若$\angle 1=\frac{1}{3}\angle ABD$，$\angle 2=\frac{1}{3}\angle ACD$，试探究$\angle BDC$，$\angle BEC$和$\angle BAC$之间的数量关系，并说明理由。
$2\angle BDC + \angle BAC = 3\angle BEC$
理由:$\because \angle 1 = \frac{1}{3}\angle ABD$,$\angle 2 = \frac{1}{3}\angle ACD$,$\therefore \angle ABE = \frac{2}{3}\angle ABD$,$\angle ACE = \frac{2}{3}\angle ACD$.$\because$由题意知,$\angle BEC = \angle BAC + \angle ABE + \angle ACE = \angle BAC + \frac{2}{3}\angle ABD + \frac{2}{3}\angle ACD$①,$\angle BDC = \angle BAC + \angle ABD + \angle ACD$②,$\therefore$②$ + $①,得$\angle BDC + \angle BEC = 2\angle BAC + \frac{5}{3}\angle ABD + \frac{5}{3}\angle ACD$.$\therefore 3\angle BDC + 3\angle BEC = 6\angle BAC + 5\angle ABD + 5\angle ACD$.$\therefore 3\angle BDC + 3\angle BEC = \angle BAC + 5(\angle BAC + \angle ABD + \angle ACD)$.$\therefore 3\angle BDC + 3\angle BEC = \angle BAC + 5\angle BDC$.$\therefore 2\angle BDC + \angle BAC = 3\angle BEC$.

答案： 典例3　
(1)$150^{\circ}$.
(2)$\angle BDC + \angle BAC = 2\angle BEC$.
理由:由题意,得$\angle BDC = \angle BEC + \angle 1 + \angle 2$①,$\angle BEC = \angle BAC + \angle ABE + \angle ACE$②.$\because BE$平分$\angle ABD$,$CE$平分$\angle ACD$,$\therefore \angle ABE = \angle 1$,$\angle ACE = \angle 2$.①$ - $②,得$\angle BDC - \angle BEC = \angle BEC - \angle BAC$.$\therefore \angle BDC + \angle BAC = 2\angle BEC$.
(3)$2\angle BDC + \angle BAC = 3\angle BEC$.
理由:$\because \angle 1 = \frac{1}{3}\angle ABD$,$\angle 2 = \frac{1}{3}\angle ACD$,$\therefore \angle ABE = \frac{2}{3}\angle ABD$,$\angle ACE = \frac{2}{3}\angle ACD$.$\because$由题意知,$\angle BEC = \angle BAC + \angle ABE + \angle ACE = \angle BAC + \frac{2}{3}\angle ABD + \frac{2}{3}\angle ACD$①,$\angle BDC = \angle BAC + \angle ABD + \angle ACD$②,$\therefore$②$ + $①,得$\angle BDC + \angle BEC = 2\angle BAC + \frac{5}{3}\angle ABD + \frac{5}{3}\angle ACD$.$\therefore 3\angle BDC + 3\angle BEC = 6\angle BAC + 5\angle ABD + 5\angle ACD$.$\therefore 3\angle BDC + 3\angle BEC = \angle BAC + 5(\angle BAC + \angle ABD + \angle ACD)$.$\therefore 3\angle BDC + 3\angle BEC = \angle BAC + 5\angle BDC$.$\therefore 2\angle BDC + \angle BAC = 3\angle BEC$.

[变式]如图①，在$\triangle ABC$中，$BD$平分$\angle ABC$，$CE$平分$\angle ACB$，$BD$与$CE$相交于点$M$。
(1)若$\angle A=50^{\circ}$，求$\angle CMD$的度数。
(2)如图②，若$MN\perp BC$于点$N$，$\angle A=60^{\circ}$，求$\angle 1-\angle 2$的度数。
(3)若$\angle BEC=x$，$\angle BDC=y$，则$\angle BMC=$______(用含$x$，$y$的式子表示)。

答案：
[变式]　
(1)$\because \angle A = 50^{\circ}$,$\therefore \angle ABC + \angle ACB = 180^{\circ} - \angle A = 130^{\circ}$.$\because BD$平分$\angle ABC$,$CE$平分$\angle ACB$,$\therefore \angle MBC = \frac{1}{2}\angle ABC$,$\angle MCB = \frac{1}{2}\angle ACB$.$\therefore \angle CMD = \angle MBC + \angle MCB = \frac{1}{2}(\angle ABC + \angle ACB) = 65^{\circ}$.
(2)$\because \angle A = 60^{\circ}$,$\therefore \angle ABC + \angle ACB = 180^{\circ} - \angle A = 120^{\circ}$.同理
(1),得$\angle MBC + \angle MCB = \frac{1}{2}(\angle ABC + \angle ACB) = 60^{\circ}$.$\therefore \angle BMC = 180^{\circ} - (\angle MBC + \angle MCB) = 120^{\circ}$.$\therefore \angle 1 + \angle BMN = 120^{\circ}$①.$\because MN \perp BC$,$\therefore \angle 2 + \angle BMN = 90^{\circ}$②.①$ - $②,得$\angle 1 - \angle 2 = 30^{\circ}$.
(3)$60^{\circ} + \frac{x + y}{3}$. 解析:如图,易得$\angle BMC = 180^{\circ} - \frac{1}{2}(\angle ABC + \angle ACB) = 180^{\circ} - \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} + \frac{1}{2}\angle A$.$\therefore \angle A = 2\angle BMC - 180^{\circ}$.$\because$易知$\angle 4 = \angle 5$,$\angle 6 = \angle 7$,$\angle BEC = \angle A + \angle 6$,$\angle BDC = \angle A + \angle 4$,$\therefore x = \angle A + \angle 7$①,$y = \angle A + \angle 5$②.①$ + $②,得$x + y = 2\angle A + (\angle 7 + \angle 5)$.$\because \angle 7 + \angle 5 = 180^{\circ} - \angle BMC$,$\therefore x + y = 4\angle BMC - 360^{\circ} + 180^{\circ} - \angle BMC$.$\therefore \angle BMC = 60^{\circ} + \frac{x + y}{3}$.