三、解答题
15.如图，$OC⊥AB$于点$O$，$OD$平分$∠BOC$，$OE$平分$∠AOD$.
(1)求$∠BOD$的度数；
(2)求$∠COE$的度数.

解:(1)$\because OC⊥AB, \therefore ∠BOC=90°. \because OD$平分$∠BOC, \therefore ∠BOD=\frac{1}{2}∠BOC=\frac{1}{2}×90°=45°$.
(2)由(1)可知$∠BOD=45°, \therefore ∠AOD=180°-45°=135°. \because OE$平分$∠AOD , \therefore ∠AOE=\frac{1}{2}∠AOD = \frac{1}{2} × 135°= 67.5°. \because OC ⊥ AB, \therefore ∠AOC=90°, \therefore ∠COE=∠AOC-∠AOE=90°-67.5°=22.5°$.

16.如图,已知∠DAB=65°,∠1=∠C.
(1)在图中画出∠DAB的对顶角;
(2)写出∠1的同位角;
(3)写出∠C的同旁内角;
(4)求∠B的度数.

解:(1)如图,∠GAH即为所求.
(2)∠1的同位角是∠DAB.
(3)∠C的同旁内角是∠B和∠ADC.
(4)$\because ∠1=∠C, \therefore AE// BC. \therefore ∠DAB+∠B=180°, \therefore ∠B=180°-∠DAB=180°-65°=115°$.

17.完成下面的解题过程,并在括号内填上依据.
如图,$∠AHF+∠FMD=180°,GH$平分$∠AHF,MN$平分$∠DME$.
求证:$GH// MN$.
证明:$\because ∠AHF+∠FMD=180°$, $\underline{\hspace{3cm}}+∠FMD=180°$,
$\therefore \underline{\hspace{6cm}}$.
$\because GH$平分$∠AHF,MN$平分$∠DME$,
$\therefore ∠1=\frac{1}{2}∠AHF,∠2=\frac{1}{2}∠DME(\underline{\hspace{6cm}})$.
$\therefore ∠1=∠2(\underline{\hspace{6cm}})$.
$\therefore GH// MN(\underline{\hspace{6cm}})$.

证明: $\because ∠AHF+∠FMD=180°, ∠DME+∠FMD=180°$,
$\therefore ∠AHF=∠DME$.
$\because GH$平分$∠AHF,MN$平分$∠DME$,
$\therefore ∠1=\frac{1}{2}∠AHF, ∠2 =\frac{1}{2}∠DME$(角平分线的定义).
$\therefore ∠1 =∠2$(等量代换).
$\therefore GH// MN$(内错角相等,两直线平行).