a: AB//CD
=>\(\dfrac{IA}{IC}=\dfrac{IB}{ID}=\dfrac{AB}{CD}=k\)
=>\(IA=k\cdot IC;IB=k\cdot ID\)
Vì IB=kID
nên \(S_{ABI}=k\cdot S_{BIC}\left(1\right)\)
Vì IA=kIC
nên \(S_{ABI}=k\cdot S_{AID}\left(2\right)\)
Từ (1),(2) suy ra \(S_{BIC}=S_{AID}\)
Kẻ AH\(\perp\)DC; BK\(\perp\)DC; CE\(\perp\)AB; DF\(\perp\)AB
=>AH//BK; CE//DF
Xét tứ giác ABKH có
AB//HK
AH//BK
Do đó: ABKH là hình bình hành
=>AH=BK
Xét tứ giác DCEF có
DC//EF
DF//CE
Do đó: DCEF là hình bình hành
=>FD=CE
\(S_{ADC}=\dfrac{1}{2}\cdot AH\cdot DC;S_{BDC}=\dfrac{1}{2}\cdot BK\cdot DC\)
mà AH=BK
nên \(S_{ADC}=S_{BDC}\)
\(S_{DAB}=\dfrac{1}{2}\cdot DF\cdot AB;S_{ABC}=\dfrac{1}{2}\cdot CE\cdot AB\)
mà DF=CE
nên \(S_{ABD}=S_{ABC}\)
b: \(S_{BIC}=S_{AID}\)
=>\(S_{AID}=10\left(cm^2\right)\)
\(\dfrac{IA}{IC}=\dfrac{S_{AIB}}{S_{BIC}}=\dfrac{4}{10}=\dfrac{2}{5}\)
=>IB/ID=2/5
=>\(\dfrac{S_{BIC}}{S_{DIC}}=\dfrac{IB}{ID}=\dfrac{2}{5}\)
=>\(S_{DIC}=25\left(cm^2\right)\)
\(S_{ABCD}=S_{AIB}+S_{BIC}+S_{AID}+S_{DIC}\)
\(=4+10+10+25=49\left(cm^2\right)\)
