$證明:∵∠3=∠4$ $∴180°-∠3=180°-∠4,即∠ACB=∠ACD$ $在△ABC和△ADC中$ ${{\begin{cases} {{∠1=∠2}} \\ {AC=AC} \\ {∠ACB=∠ACD} \end{cases}}}$ $∴△ABC≌△ADC(ASA)$ $∴AB=AD$
$證明:∵AD//BE$ $∴∠A=∠EBC$ $∵BD//CE$ $∴∠DBA=∠C$ $∵B是AC中點$ $∴AB=BC $ $在△ABD和△BCE中$ ${{\begin{cases} {{∠A=∠CBE}} \\ {AB=BC} \\ {∠ABD=∠C} \end{cases}}}$ $∴△ABD≌△BCE(ASA)$
$證明:∵∠ADE=∠2+∠BDE=∠1+∠C$ $又∵∠1=∠2$ $∴∠BDE=∠C$ $在△AEC和△BED中$ ${{\begin{cases} {{∠C=∠BDE}} \\ {∠A=∠B} \\ {AE=BE} \end{cases}}}$ $∴△AEC≌△BED(AAS)$ $∴EC=ED$
$解:延長AP交BC于D$
$∵在△ABD中,BP既是角平分線又是高$ $∴BA=BD,PA=PD$ $∴S_{△BPD}=\frac {1}{2}S_{△ABD},S_{△CPD}=\frac {1}{2}S_{△CAD}$ $∴S_{△PBC}=S_{△PBD}+S_{△PCD}=\frac {1}{2}S_{△ABC}$ $=8 cm^{2}$
$證明:∵AC//BG$ $∴∠C=∠GBD$ $∵D是BC中點$ $∴BD=CD$ $在△BDG和△CDF中$ ${{\begin{cases} {{∠BDG=∠CDF}} \\ {BD=CD} \\ {∠GBD=∠C} \end{cases}}}$ $∴△BDG≌△CDF(ASA)$ $∴BG=CF$
解:BE+CF>EF,理由如下: 由(1)知,△BDG≌△CDF ∴DG=DF ∴ED既是△EGF的中線又是高 ∴△EGF是等腰三角形,EG=EF 在△BEG中,BE+BG>EG 即BE+CF>EF
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