本帖最后由 郝酒 于 2018-7-29 20:37 编辑
看到一个解答,简述如下.
连PB,PD,PC,PE,可知$\angle PEC=\angle PDB,\angle PCA = \angle PBA$ 所以$\triangle PBD \sim \triangle PCE$,进而$\angle BPD = \angle CPE,\angle BPC = \angle DPE$,$\frac{PB}{PC}=\frac{PD}{PE}$,所以$\triangle PBC \sim \triangle PDE$
又因为$\angle BMC = 180^\circ -\angle A=\angle B+\angle C = 2\angle DIE=\angle DOE$,所以$\triangle BMC \sim \triangle DOE$
所以$\frac{IM}{IO}=\frac{BM}{DO}=\frac{BC}{DE}=\frac{PC}{PE}$
最后证明$\angle OIM = \angle EPC$
$\angle EPC=\angle (DE,BC)=\angle AED - \angle ACB$
$\angle OIM = \angle DIM - \angle DIO=\frac{1}{2}\angle B-\frac{1}{2}(180^\circ - 2\angle IED)=\frac{1}{2}\angle B+\angle IED-90^\circ=\frac{1}{2}\angle B + \angle AED - \angle AEI-90^\circ=\angle AED +\frac{1}{2}\angle B -\frac{1}{2}(\angle C - \angle A)-90^\circ =\angle AED - \angle ACB$
得$\triangle IOM \sim \triangle PEC$,$\angle IMP = \angle ACP = \angle CPE + \angle CEP = \angle MIO+\angle MOI$,得证. |