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17. (10分)如图，在△ABC中，作AC边的垂直平分线MN交BC于点D，连接AD.
(1)根据题意补全图形. (保留作图痕迹，不写作法)
(2)若AB = AD，∠BAD = 40°，求∠C的度数.
(1)根据题意补全图形. (保留作图痕迹，不写作法)
(2)若AB = AD，∠BAD = 40°，求∠C的度数.
答案：
17.
(1)如图，即为所求.
(2)解：$\because AB = AD$，$\angle BAD = 40^{\circ}$，
$\therefore \angle B = \angle 1 = \frac{1}{2}(180^{\circ} - \angle BAD) = 70^{\circ}$.
$\because MN$垂直平分AC，$\therefore AD = CD$.$\therefore \angle C = \angle 2$.
$\because \angle 1 = \angle C + \angle 2$，$\therefore \angle C = \frac{1}{2} \angle 1 = 35^{\circ}$.
17.
(1)如图，即为所求.
(2)解：$\because AB = AD$，$\angle BAD = 40^{\circ}$，
$\therefore \angle B = \angle 1 = \frac{1}{2}(180^{\circ} - \angle BAD) = 70^{\circ}$.
$\because MN$垂直平分AC，$\therefore AD = CD$.$\therefore \angle C = \angle 2$.
$\because \angle 1 = \angle C + \angle 2$，$\therefore \angle C = \frac{1}{2} \angle 1 = 35^{\circ}$.
18. (8分)如图，已知△ABC三个顶点的坐标分别是点A(-5,0)，B(-2,4)，C(-1,-2)，△A'B'C'与△ABC关于y轴对称.
(1)直接写出点A'，B'，C'的坐标.
(2)在图中画出△A'B'C'.
(3)求△ABC的面积.
(1)直接写出点A'，B'，C'的坐标.
(2)在图中画出△A'B'C'.
(3)求△ABC的面积.
答案：
18.
(1)点$A'(5,0)$，$B'(2,4)$，$C'(1,-2)$.
(2)
(3)解：$S_{\triangle ABC} = 4 × 6 - \frac{1}{2} × 3 × 4 - \frac{1}{2} × 2 × 4 - \frac{1}{2} × 1 × 6 = 11$.
$\therefore \triangle ABC$的面积是11.
18.
(1)点$A'(5,0)$，$B'(2,4)$，$C'(1,-2)$.
(2)
(3)解：$S_{\triangle ABC} = 4 × 6 - \frac{1}{2} × 3 × 4 - \frac{1}{2} × 2 × 4 - \frac{1}{2} × 1 × 6 = 11$.
$\therefore \triangle ABC$的面积是11.
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