(Bạn xem bài 1, 3b)
Bài 1:
Kẻ AH $\bot$ BC, AH $\cap$ MN= {I}
$S_{ABC}= \frac{1}{2}.AH.BC $
Có MN= $\frac{1}{2}$.BC (đường trung bình)
$\Delta$ ABH có MI//BH
=> $\frac{AI}{AH}= \frac{AM}{AB}= \frac{1}{2}$
=> AI= $\frac{1}{2}$.AH
$S_{AMN}= \frac{1}{2}.AI.MN$
$= \frac{1}{2}.\frac{1}{2}.AH.\frac{1}{2}.BC$
$= \frac{1}{4}.S_{ABC}$
=> $S_{MNCB}= \frac{3}{4}.S_{ABC}$
Ta có $\Delta$ EBM= $\Delta$ IAM (ch.gn)
=> $S_{MBE}= S_{MAI}$
Tương tự, $S_{AIN}= S_{CFN}$
=> $S_{MBE}+ S_{CFN}= S_{AMI}+ S_{AIN}= S_{AMN}= \frac{1}{4}S_{ABC}$
=> $S_{BEFC}= S_{MBCN}+ S_{EBM}+ S_{NCF}= \frac{3}{4}.S_{ABC}+ \frac{1}{4}.S_{ABC}= S_{ABC}$ (đpcm)
Bài 3:
a, Thiếu dữ kiện
b,
$S_{DEK}= \frac{1}{2}.ab$
DC= DK-CK= a-b
=> $S_{ABCD}= (a-b)^2$
$S_{AEB}= \frac{1}{2}.(a-b).b$
$S_{ABKD}= \frac{(a-b+a).(a-b)}{2}= \frac{(2a-b)(a-b)}{2}$
=> $S_{BEK}= \frac{(2a-b)(a-b)}{2} - \frac{b(a-b)}{2} - \frac{2(a-b)}{2}$
$= \frac{(2a-b)(a-b)-b(a-b)-2(a-b)}{2}$
$= \frac{(a-b)(2a-b-b-2)}{2}$
$= \frac{(a-b)(2a-2b-2)}{2}$