《2025年思维新观察八年级数学上册人教版湖北专版》

9.(2025·黄冈)如图,$BP是\angle ABC$的平分线,$AP\perp BP于点P$,连接$PC$,若$\triangle ABC的面积为1\mathrm{c}{\mathrm{m}}^{\mathrm{2}}$,则$\triangle PBC$的面积为(B)

A.$0.4\mathrm{c}{\mathrm{m}}^{\mathrm{2}}$
B.$0.5\mathrm{c}{\mathrm{m}}^{\mathrm{2}}$
C.$0.6\mathrm{c}{\mathrm{m}}^{\mathrm{2}}$
D.不能确定

答案： B
提示:延长AP交BC于点M,
$\therefore \triangle ABP\cong \triangle MBP(ASA)$,
$\therefore S_{\triangle BPC}=\frac {1}{2}S_{\triangle ABC}=0.5cm^{2}$

10.如图,在$\triangle ABC$中,$AC\perp BC$,$CE\perp AB于E$,$AF平分\angle CAB交CE于点F$,过$F作FD// BC交AB于点D$.
(1)求证:$\angle ADF = \angle ACE$;
(2)求证:$AC = AD$.

答案：
证明:
(1)$\because ∠1+∠3=∠2+∠3=90^{\circ }$,$\therefore ∠1=∠2$,
$\because DF// BC$,$\therefore ∠1=∠2=∠4$,
$\therefore ∠ADF=∠ACE$;

(2)$\because AF$平分$∠CAB$,$\therefore ∠5=∠6$,
$\triangle ACF\cong \triangle ADF(AAS)$,$\therefore AC=AD$.

11.如图,$\triangle ABC$中,点$P在AB$上,点$M$,$N分别在AC$,$BC$上,$\angle A = \angle B = \angle MPN = 60^{\circ}$,$PA = BN$,求证:$PM = PN$.
证明:∠MPB=∠A+∠AMP=60°+∠BPN,
$\therefore$∠BPN=∠AMP,
$\therefore$△APM≌△BNP(AAS),
$\therefore PM=PN$.

答案： 证明:$∠MPB=∠A+∠AMP=60^{\circ }+∠BPN$,
$\therefore ∠BPN=∠AMP$,
$\therefore \triangle APM\cong \triangle BNP(AAS)$,
$\therefore PM=PN$.

12.利用无刻度直尺完成下列作图,$A$,$B$都为格点.
(1)在图1中作$AB的中点D$; (2)在图2中作直线$BE平分\triangle ABC$面积;
(3)在图3中,过$C点作CM\perp AB$; (4)如图4,点$C$在格线上,作$BC边上的中线AD$.

答案：
如图所示.

13.如图,在$\mathrm{R}\mathrm{t}\triangle ABC$中,$AC = BC$,$AC\perp BC$,$BF平分\angle ABC$,交$AC于点F$,$CM\perp BF于点E$,交$AB于点M$,$AD\perp CM交CM的延长线于D$点.
(1)求证:$\triangle ACD\cong\triangle CBE$; (2)求证:$BF = CD + DM$.

证明:(1)$\triangle ACD\cong \triangle CBE$;
(2)$AD=CE$.
又$∠DAM=∠ABF=∠FBC=∠ACD$,
$∠ADM=∠CEF=90^{\circ }$,
$\therefore \triangle ADM\cong \triangle CEF$,
$\therefore EF=DM$,
$\therefore BF=CD+EF=CD+DM$.

答案： 证明:
(1)$\triangle ACD\cong \triangle CBE$;
(2)$AD=CE$.
又$∠DAM=∠ABF=∠FBC=∠ACD$,
$∠ADM=∠CEF=90^{\circ }$,
$\therefore \triangle ADM\cong \triangle CEF$,
$\therefore EF=DM$,
$\therefore BF=CD+EF=CD+DM$.