《2025年思维新观察八年级数学上册人教版湖北专版》

问题:在等边$\triangle ABC$中,$D$,$E分别在BC$,$AB$上,连接$AD$,$CE交于点M$,$BE = CD$.连接$BM$,若$BM\perp AM$,求证:$AM = 2CM$.

答案： 证明:方法一:作 $ AN \perp CE $,垂足为 $ N $,证 $ \triangle EBC \cong \triangle DCA $,证 $ \angle BAM = \angle ACN $,证明 $ \triangle CAN \cong \triangle ABM $,$ \therefore AM = CN $,又 $ AM = 2MN $,$ \therefore CM = MN $,$ \therefore AM = 2CM $.方法二:在 $ AM $ 上取点 $ G $,使 $ AG = CM $,连接 $ CG $,$ \therefore \triangle ACG \cong \triangle CBM $,$ \therefore \angle AGC = \angle BMC = 150^{\circ} $,$ MG = CM $,$ \therefore AM = 2CM $.

变式1.如图,$\triangle ABC$,$\triangle DPC$都是等边三角形.点$P在\triangle ABC$内,$M为AC$的中点,连接$PM$,$PA$,$PB$,若$PA\perp PM$,且$PB = 2PM$.
(1)求证:$BP\perp BD$;
证明:$\triangle CAP \cong \triangle CBD \Rightarrow AP = BD $,$ \angle BDC = 120^{\circ} $,$ \therefore \angle BDP = 60^{\circ} $.延长 $ PM $ 至 $ N $ 使 $ MN = PM $,连接 $ AN $,$ CN $,$ \therefore \triangle APM \cong \triangle CNM $,$ \therefore \angle PNC = 90^{\circ} $,$ CN = AP $,$ \angle PCN = 60^{\circ} $,$ \therefore \triangle PCN \cong \triangle PDB(SAS) $,$ \therefore \angle PBD = \angle PNC = 90^{\circ} $,$ \therefore BP \perp BD $
(2)判断$PC与PA$的数量关系并证明.
$PC = 2PA$,证明:在 $ \triangle PBD $ 中,$ \angle BPD = 30^{\circ} $,$ PD = 2BD = 2PA $,$ \therefore PC = 2PA $.

答案：
(1)证明:$ \triangle CAP \cong \triangle CBD \Rightarrow AP = BD $,$ \angle BDC = 120^{\circ} $,$ \therefore \angle BDP = 60^{\circ} $.延长 $ PM $ 至 $ N $ 使 $ MN = PM $,连接 $ AN $,$ CN $,$ \therefore \triangle APM \cong \triangle CNM $,$ \therefore \angle PNC = 90^{\circ} $,$ CN = AP $,$ \angle PCN = 60^{\circ} $,$ \therefore \triangle PCN \cong \triangle PDB(SAS) $,$ \therefore \angle PBD = \angle PNC = 90^{\circ} $,$ \therefore BP \perp BD $;
(2)在 $ \triangle PBD $ 中,$ \angle BPD = 30^{\circ} $,$ PD = 2BD = 2PA $,$ \therefore PC = 2PA $.

变式2.如图,在平面直角坐标系中,$AB\perp BC$,$AB = BC$,$A$,$B分别在x$轴负半轴,$y$轴负半轴上,连接$AC$,$\triangle ACD$为等边三角形,$\angle AOD = 60^{\circ}$.若$B(0,-1)$,求$OC$的长.
解:在 $ OD $ 上取点 $ M $ 使 $ OM = OA $,连接 $ AM $,证 $ \triangle ADM \cong \triangle ACO $,$ \therefore \angle AOC = $120°.作 $ CH \perp y $ 轴于 $ H $ 点,$ \therefore CH = \frac{1}{2}OC $.又 $ \because \triangle AOB \cong \triangle BHC $,$ \therefore CH = $1,$ \therefore OC = $2.

答案： 解:在 $ OD $ 上取点 $ M $ 使 $ OM = OA $,连接 $ AM $,证 $ \triangle ADM \cong \triangle ACO $,$ \therefore \angle AOC = 120^{\circ} $.作 $ CH \perp y $ 轴于 $ H $ 点,$ \therefore CH = \frac{1}{2}OC $.又 $ \because \triangle AOB \cong \triangle BHC $,$ \therefore CH = 1 $,$ \therefore OC = 2 $.