4. 已知直线$AB// CD$,$P$为平面内一点,连接$PA$,$PD$.
(1)如图①,已知$∠ A=50°$,$∠ D=150°$,求$∠ APD$的度数;
(2)如图②,判断$∠ PAB$,$∠ CDP$,$∠ APD$之间的数量关系,并说明理由;
(3)如图③,$AP⊥ PD$,$DN$平分$∠ PDC$,$AN$交$DP$于点$O$,$∠ PAN+\frac{1}{2}∠ PAB=∠ APD$,求$∠ AND$的度数.
答案：4.解:(1)如答图①,过点P作$PE// AB.$
$\because AB// CD,\therefore AB// PE// CD,$
$\therefore ∠APE=∠A,∠CDP+∠EPD=180° .$
$\because ∠A=50° ,∠D=150° ,$
$\therefore ∠APE=50° ,∠EPD=180° -150° =30° ,$
$\therefore ∠APD=∠APE+∠EPD=50° +30° =80° .$
(2)$∠CDP+∠PAB-∠APD=180°$.理由如下:
如答图②,过点P作$PF// AB$,则$AB// PF// CD,$
$\therefore ∠CDP=∠DPF,∠FPA+∠PAB=180° ,$
$\because ∠FPA=∠DPF-∠APD,$
$\therefore ∠DPF-∠APD+∠PAB=180° ,$
$\therefore ∠CDP+∠PAB-∠APD=180° .$
(3)$\because AP⊥PD,\therefore ∠APO=90° .$
$\because ∠PAN+\frac{1}{2}∠PAB=∠APD,$
$\therefore ∠PAN+\frac{1}{2}∠PAB=90° .$
$\because ∠POA+∠PAN=90° ,\therefore ∠POA=\frac{1}{2}∠PAB.$
$\because ∠POA=∠NOD,\therefore ∠NOD=\frac{1}{2}∠PAB.$
$\because DN$平分$∠PDC,\therefore ∠ODN=\frac{1}{2}∠PDC,$
$\therefore ∠AND=180° -∠NOD-∠ODN=180° -\frac{1}{2}(∠PAB+$
$∠PDC).$
由(2)得$∠CDP+∠PAB-∠APD=180° ,$
$\therefore ∠CDP+∠PAB=180° +∠APD,$
$\therefore ∠AND=180° -\frac{1}{2}(∠PAB+∠PDC)=180° -$
$\frac{1}{2}(180° +∠APD)=180° -\frac{1}{2}×(180° +90° )=45° .$
