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1. 如下图，AB // CD，AE交CD于点C，DE ⊥ AE，垂足为E，∠A=37°，求∠D的度数. 
答案
1. 解:$\because AB// CD,∠ A=37°$,
$\therefore ∠ ECD=∠ A=37°$.
$\because DE⊥ AE$,
$\therefore ∠ D=90°-∠ ECD=90°-37°=53°$.
$\therefore ∠ ECD=∠ A=37°$.
$\because DE⊥ AE$,
$\therefore ∠ D=90°-∠ ECD=90°-37°=53°$.
2. 如下图,直线AD与AB,CD分别相交于点A,D,EC,BF分别与AB,CD相交于E,C,B,F.如果∠1=∠2,∠B=∠C,求证:∠A=∠D.
答案
2. 证明:$\because ∠ 1=∠ 2$, 且$∠ 2=∠ AGB$,
$\therefore ∠ 1=∠ AGB$,
$\therefore EC// BF$,
$\therefore ∠ C=∠ BFD$.
又$\because ∠ B=∠ C$,
$\therefore ∠ B=∠ BFD$,
$\therefore AB// CD$,
$\therefore ∠ A=∠ D$.
$\therefore ∠ 1=∠ AGB$,
$\therefore EC// BF$,
$\therefore ∠ C=∠ BFD$.
又$\because ∠ B=∠ C$,
$\therefore ∠ B=∠ BFD$,
$\therefore AB// CD$,
$\therefore ∠ A=∠ D$.
3.如下图，AB⊥CD，垂足为O，EF经过点O，∠2=4∠1，求∠2，∠3，∠BOE的度数.
答案
3. 解:$\because ∠ 2=4∠ 1,∠ 1+∠ 2=90°(AB⊥ CD)$,
$\therefore ∠ 1=90°÷ 5=18°$.
$\therefore ∠ 2=4∠ 1=4×18°=72°,∠ 3=∠ 1=18°$.
$\therefore ∠ BOE=180°-∠ 1=180°-18°=162°$.
$\therefore ∠ 1=90°÷ 5=18°$.
$\therefore ∠ 2=4∠ 1=4×18°=72°,∠ 3=∠ 1=18°$.
$\therefore ∠ BOE=180°-∠ 1=180°-18°=162°$.
