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$3$

$6:5:3$

 证明：$\because PB\perp AB，$$PC\perp AC，$$PB = PC，$
$\therefore AP$平分$\angle BAC，$即$\angle BAP = \angle CAP.$

 $\because \angle BAP + \angle BPA = 90^{\circ}，$$\angle CAP + \angle CPA = 90^{\circ}，$
$\therefore \angle BPA = \angle CPA.$

 在$\triangle PBD$和$\triangle PCD$中，

$\begin{cases}PB = PC,\\\angle BPD = \angle CPD,\\PD = PD,\end{cases}$

 $\therefore \triangle PBD\cong\triangle PCD(\text{SAS})，$
$\therefore \angle BDP = \angle CDP.$

 (1) 证明：如图，过点$P$作$PF\perp BE$于点$F，$$PN\perp BD$于点$N，$
$PM\perp AC$于点$M.$

 $\because CP$平分$\angle ACD，$$\therefore PM = PN.$

 $\because BP$平分$\angle ABC，$$\therefore PF = PN，$
$\therefore PF = PM.$

 又$\because PF\perp AE，$$PM\perp AC，$
$\therefore AP$平分$\angle CAE.$

 (2) 设$\angle ACP = \angle PCD = x.$ 由

 (1)知$AP$平分$\angle CAE，$$\therefore \angle FAP = \angle PAC.$

 $\because \angle BPC = 40^{\circ}，$$\therefore \angle ABP = \angle PBC = x - 40^{\circ}.$

 $\therefore \angle BAC = \angle ACD - \angle ABC = 2x-(x - 40^{\circ})-(x - 40^{\circ}) = 80^{\circ}.$

 $\therefore \angle CAF = 180^{\circ}-\angle BAC = 100^{\circ}，$$\therefore \angle CAP=\frac{1}{2}\angle CAF = 50^{\circ}.$

 解：(1) $\because EF\perp AB，$$\angle AEF = 50^{\circ}，$
$\therefore \angle FAE = 90^{\circ}-50^{\circ}=40^{\circ}.$

 $\because \angle BAD = 100^{\circ}，$
$\therefore \angle CAD = 180^{\circ}-100^{\circ}-40^{\circ}=40^{\circ}.$

 (2) 过点$E$作$EG\perp AD$于点$G，$$EH\perp BC$于点$H.$

 $\because \angle FAE = \angle DAE = 40^{\circ}，$$EF\perp BF，$$EG\perp AD，$
$\therefore EF = EG.$

 $\because BE$平分$\angle ABC，$$EF\perp BF，$$EH\perp BC，$
$\therefore EF = EH，$$\therefore EH = EG.$

 $\because EG\perp AD，$$EH\perp BC，$
$\therefore DE$平分$\angle ADC.$

 (3) $\because S_{\triangle ACD}=15，$
$\therefore \frac{1}{2}AD\cdot EG+\frac{1}{2}CD\cdot EH = 15，$

即$\frac{1}{2}\times4\times EG+\frac{1}{2}\times8\times EG = 15.$

 $\therefore EG = EH=\frac{5}{2}，$$\therefore EF = EH=\frac{5}{2}.$

 $\therefore S_{\triangle ABE}=\frac{1}{2}AB\cdot EF=\frac{1}{2}\times7\times\frac{5}{2}=\frac{35}{4}.$