a: XétΔABC có AD là phân giác
nên \(\frac{DB}{DC}=\frac{AB}{AC}=\frac{12}{18}=\frac23\)
=>\(BD=\frac25BC;CD=\frac35BC\)
AI+ID=AD
=>AD=2ID+ID=3ID
=>\(AI=\frac23AD\)
Ta có: \(BD=\frac25BC\)
=>\(S_{ABD}=\frac25\cdot S_{ABC}\)
Vì \(AI=\frac23AD\) nên \(S_{ABI}=\frac23\cdot S_{ABD}=\frac23\cdot\frac25\cdot S_{ABC}=\frac{4}{15}\cdot S_{ABC}\)
Ta có; \(CD=\frac35\cdot BC\)
=>\(S_{CDA}=\frac35\cdot S_{CAB}\)
TA có: \(AI=\frac23AD\)
=>\(S_{AIC}=\frac23\cdot S_{ADC}=\frac23\cdot\frac35\cdot S_{ABC}=\frac25\cdot S_{ABC}\)
Ta có: \(S_{AIB}+S_{AIC}+S_{BIC}=S_{ABC}\)
=>\(S_{BIC}=S_{ABC}-\frac25\cdot S_{ABC}-\frac{4}{15}\cdot S_{ABC}=S_{ABC}\left(\frac35-\frac{4}{15}\right)=S_{ABC}\cdot\frac13\)
Ta có: \(\frac{S_{BIA}}{S_{BIC}}=\frac{4}{15}:\frac13=\frac{4}{15}\cdot3=\frac45\)
E nằm giữa A và C
=>\(\frac{S_{BEA}}{S_{BEC}}=\frac{EA}{EC};\frac{S_{IEA}}{S_{IEC}}=\frac{EA}{EC}\)
=>\(\frac{S_{BEA}-S_{IEA}}{S_{BEC}-S_{IEC}}=\frac{EA}{EC}\)
=>\(\frac{S_{BIA}}{S_{BIC}}=\frac{EA}{EC}\)
=>\(\frac{EA}{EC}=\frac45\)
b: \(\frac{EA}{EC}=\frac45\)
=>\(EA=\frac45\cdot EC\)
EA+EC=AC
=>\(\frac45\cdot EC+EC=AC\)
=>1,8EC=AC
=>1,8EC=18
=>EC=10(cm)
=>\(EA=10\cdot\frac45=8\left(\operatorname{cm}\right)\)
