9. 如图，点$D$在$BC$的延长线上，$DE\perp AB$于点$E$，交$AC$于点$F$. 若$\angle A = 35^{\circ}$，$\angle D = 15^{\circ}$，则$\angle ACB$的度数为（ ）
A. $65^{\circ}$
B. $70^{\circ}$
C. $75^{\circ}$
D. $85^{\circ}$
A. $65^{\circ}$
B. $70^{\circ}$
C. $75^{\circ}$
D. $85^{\circ}$
答案
B
10. 如图，$\angle 1$，$\angle 2$，$\angle 3$，$\angle 4$的关系正确的是（ ）
A. $\angle 1 + \angle 2 = \angle 3 + \angle 4$
B. $\angle 1 - \angle 3 = \angle 2 - \angle 4$
C. $\angle 1 + \angle 2 = \angle 4 - \angle 3$
D. $\angle 1 + \angle 3 = \angle 2 + \angle 4$
A. $\angle 1 + \angle 2 = \angle 3 + \angle 4$
B. $\angle 1 - \angle 3 = \angle 2 - \angle 4$
C. $\angle 1 + \angle 2 = \angle 4 - \angle 3$
D. $\angle 1 + \angle 3 = \angle 2 + \angle 4$
答案
C
11. 在$\triangle ABC$中，三个内角$\angle A$，$\angle B$，$\angle C$满足$\angle B - \angle A = \angle C - \angle B$，则$\angle B$的度数为_______.
答案
$60^{\circ}$ 解析：由$\angle B - \angle A=\angle C - \angle B$，得$2\angle B=\angle A+\angle C$. 根据三角形的内角和是$180^{\circ}$，得$\angle A+\angle B+\angle C = 180^{\circ}$，所以$3\angle B = 180^{\circ}$，解得$\angle B = 60^{\circ}$.
12. 如图，在$\triangle ABC$中，$AD$平分$\angle BAC$. 若$\angle 2 = 70^{\circ}$，则$\angle 1 + \angle 3 =$_______$^{\circ}$.
答案
140 解析：$\because AD$平分$\angle BAC$，$\therefore\angle BAD=\angle CAD$.
$\because\angle 2=\angle 1+\angle BAD$，$\angle 3=\angle 2+\angle CAD$，$\therefore\angle 1+\angle BAD+\angle 3=\angle 2+\angle 2+\angle CAD$，$\therefore\angle 1+\angle 3 = 2\angle 2$. 又$\because\angle 2 = 70^{\circ}$，$\therefore\angle 1+\angle 3 = 2\times70^{\circ}=140^{\circ}$.
$\because\angle 2=\angle 1+\angle BAD$，$\angle 3=\angle 2+\angle CAD$，$\therefore\angle 1+\angle BAD+\angle 3=\angle 2+\angle 2+\angle CAD$，$\therefore\angle 1+\angle 3 = 2\angle 2$. 又$\because\angle 2 = 70^{\circ}$，$\therefore\angle 1+\angle 3 = 2\times70^{\circ}=140^{\circ}$.
13. 如图，在$\triangle ABC$中，$BO$，$CO$分别平分$\angle ABC$和$\angle ACB$.
（1）若$\angle A = 40^{\circ}$，则$\angle BOC$的度数为_______；
（2）求证：$\angle BOC = 90^{\circ}+\frac{1}{2}\angle A$.
（1）若$\angle A = 40^{\circ}$，则$\angle BOC$的度数为_______；
（2）求证：$\angle BOC = 90^{\circ}+\frac{1}{2}\angle A$.
答案
(1) $110^{\circ}$ (2) $\because\triangle ABC$的内角和为$180^{\circ}$，$\therefore\angle ABC+\angle ACB = 180^{\circ}-\angle A$. $\because BO$，$CO$分别平分$\angle ABC$和$\angle ACB$，
$\therefore\angle OBC=\frac{1}{2}\angle ABC$，$\angle OCB=\frac{1}{2}\angle ACB$，$\therefore\angle OBC+\angle OCB=\frac{1}{2}(\angle ABC+\angle ACB)=90^{\circ}-\frac{1}{2}\angle A$. $\because\triangle OBC$的内角和为$180^{\circ}$，$\therefore\angle BOC = 180^{\circ}-(\angle OBC+\angle OCB)=180^{\circ}-(90^{\circ}-\frac{1}{2}\angle A)=90^{\circ}+\frac{1}{2}\angle A$
$\therefore\angle OBC=\frac{1}{2}\angle ABC$，$\angle OCB=\frac{1}{2}\angle ACB$，$\therefore\angle OBC+\angle OCB=\frac{1}{2}(\angle ABC+\angle ACB)=90^{\circ}-\frac{1}{2}\angle A$. $\because\triangle OBC$的内角和为$180^{\circ}$，$\therefore\angle BOC = 180^{\circ}-(\angle OBC+\angle OCB)=180^{\circ}-(90^{\circ}-\frac{1}{2}\angle A)=90^{\circ}+\frac{1}{2}\angle A$
14. 如图，在$\triangle ABC$中，$AD$是$\triangle ABC$的高，$AE$平分$\angle BAC$（$\angle C＞\angle B$）.
（1）若$\angle B = 30^{\circ}$，$\angle C = 50^{\circ}$，则$\angle DAE$的度数为_______；
（2）试猜想$\angle DAE$与$\angle C - \angle B$之间的数量关系，并说明理由.
（1）若$\angle B = 30^{\circ}$，$\angle C = 50^{\circ}$，则$\angle DAE$的度数为_______；
（2）试猜想$\angle DAE$与$\angle C - \angle B$之间的数量关系，并说明理由.
答案
(1) $10^{\circ}$ (2) $\angle DAE=\frac{1}{2}(\angle C - \angle B)$ 理由：设$\angle B = x$，$\angle C = y$. 在$\triangle ABC$中，$\angle BAC = 180^{\circ}-x - y$. $\because AE$平分$\angle BAC$，$\therefore\angle BAE=\frac{1}{2}\angle BAC=\frac{1}{2}(180^{\circ}-x - y)$. $\because AD$是$\triangle ABC$的高，$\therefore\angle ADB = 90^{\circ}$，$\therefore$在$\triangle ABD$中，$\angle BAD = 180^{\circ}-\angle ADB-\angle B = 90^{\circ}-x$，$\therefore\angle DAE=\angle BAD-\angle BAE=90^{\circ}-x-\frac{1}{2}(180^{\circ}-x - y)=\frac{1}{2}(y - x)=\frac{1}{2}(\angle C - \angle B)$.
