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第111页

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线段PC的长度

$PH<PC<OC$

 解：

 $\because OE⊥ CD，$$OF⊥ AB，$

 $\therefore ∠ DOE=90°，$$∠ BOF=90°，$

 $\because ∠ EOF=145°，$

 $\therefore ∠ DOF=∠ EOF-∠ DOE=145°-90°=55°，$

 又$\because ∠ BOF=∠ BOD+∠ DOF=90°，$

 $\therefore ∠ BOD=90°-∠ DOF=90°-55°=35°。$

 垂线段最短

 解：

 $\because OE ⊥ AB$

 $\therefore ∠ AOE = 90°$

 $\because ∠ COE = 30°$

 $\therefore ∠ AOC = ∠ AOE - ∠ COE = 90° - 30° = 60°$

 又$\because$ 直线$CD$为平角，$∠ COD=180°$

 $\therefore ∠ DOA = 180° - ∠ AOC = 180° - 60° = 120°$

 解：

 (1) $OA ⊥ OC，$理由如下：

 $\because OB ⊥ OD，$

 $\therefore ∠ BOD = 90°，$即 $∠ 2 + ∠ BOC = 90°。$

 又$\because ∠ 1 = ∠ 2，$

 $\therefore ∠ 1 + ∠ BOC = 90°，$即 $∠ AOC = 90°，$

 $\therefore OA ⊥ OC。$

 (2) $\because ∠ BOC = x°，$$∠ BOD = 90°，$

 $\therefore ∠ 2 = 90° - x°，$

 又$\because ∠ 1 = ∠ 2，$

 $\therefore ∠ 1 = 90° - x°，$

 $\therefore ∠ AOD = ∠ 1 + ∠ BOD = (90° - x°) + 90° = 180° - x°。$