國中數數2題 請幫解?

1)
因 △EFC = △DFC 及 △GFB = △AFB, 故五邊形 BCEFG 面積 = △AFD = 21(2+3+4)/3 = 63.

1)
設 ABCD 面積為 1, 則 △DEF = ½(DF/DA)(DE/DC) = ½(5/9)(2/5) = 1/9
△CEH = ½ (EC/DC)(CH/CB) = ½(3/5)(1/3) = 1/10
△EFH = DCHF - △DEF - △CEH = △DFH + △DCH - △DEF - △CEH = ½(5/9) + ½(1/3) - 1/9 - 1/10 = 7/30
△BFH = ½(2/3) = 1/3
故 EG : GB = △EFH : △BFH = 7/30 : 1/3 = 7 : 10.