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16.(8分)如图，$AB//DE$，$BC = AD$，$\angle B = \angle DAE$，求证：$AC = ED.$
答案：
16.证明：$\because AB// DE,\therefore \angle E=\angle BAC$，在$\triangle ABC$与$\triangle EAD$中，$\begin{cases}\angle B=\angle DAE,\\ \angle BAC=\angle E,\\BC=AD,\end{cases}$ $\therefore \triangle ABC\cong\triangle EAD(AAS).\therefore AC=ED.$
17.(6分)如图，在$\bigtriangleup ABC$中，$BD$是$AC$边上的高，$\angle A = 70^{\circ}$，$CE$平分$\angle ACB$交$BD$于点$E$，$\angle BEC = 118^{\circ}$，求$\angle ABC$的度数.
答案：
17.解：$\because BD\perp AC,\therefore \angle BDA=\angle BDC=90^{\circ}$. $\because \angle A=70^{\circ},\therefore \angle ABD=90^{\circ}-\angle A=20^{\circ}$. $\because \angle BEC=\angle BDC+\angle DCE=118^{\circ}$， $\therefore \angle DCE=\angle BEC-\angle BDC=118^{\circ}-90^{\circ}=28^{\circ}$. $\because CE$平分$\angle ACB,\therefore \angle DCB=2\angle DCE=56^{\circ}$. $\therefore \angle DBC=90^{\circ}-\angle DCB=34^{\circ}$. $\therefore \angle ABC=\angle ABD+\angle DBC=20^{\circ}+34^{\circ}=54^{\circ}$.
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