证明:过E作EF⊥AB于F ∵BE平分∠ABC,EC⊥BC,EF⊥AB ∴EC=EF ∵E是CD中点 ∴EC=ED ∴EF=ED 又∵EF⊥AB,ED⊥AD ∴AE平分∠BAD $解:∵AC=AE,DC⊥AC,DE⊥AE$ $∴AD平分∠CAB$ $∴∠DAE=∠DAC=\frac {1}{2}∠CAE=\frac {1}{2}(90°-∠B)=\frac {1}{2}∠BDE=23°$
$ 证明:连接AM,CM$ $∵MN垂直平分AC$ $∴MA=MC$ $在Rt△ADM和Rt△CEM$ ${{\begin{cases} {{AM=CM}} \\ {AD=CE} \end{cases}}}$ $∴Rt△ADM≌Rt△CEM (HL)$ $∴MD=ME$ $又∵MD⊥BD,ME⊥BC$ $∴BM平分∠ABC$
$证明:过D作DM⊥AB于M,交PE于G$ $过D作DN⊥AC于N,交PF于H$ $又∵PE//AB$ $∴DG⊥PE,∠DPE=∠DAB$ $∵∠PFD=∠C$ $∴PF//AC$ $∴DH⊥PF,∠DPF=∠DAC$ $∵DM=DN,DM⊥AB,DN⊥AC$ $∴AD平分∠BAC,即∠BAD=∠CAD$ $∴∠DPE=∠DPF,即PD平分∠EPF$ $又∵DG⊥PE,DH⊥PF$ $∴DG=DH,即D到PE,PF距离相等$
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