Giải thích các bước giải:

a.Ta có: $S_{ABCD}=12^2=144$

b.Ta có:

$S_{ABE}=\dfrac13S_{ABCD}$

$\to S_{ABE}=\dfrac13\cdot 2S_{ABD}$

$\to S_{ABE}=\dfrac23S_{ABD}$

$\to AE=\dfrac23AD$

$\to AE=8$

c.Ta có $AE=\dfrac23AD$

$\to \dfrac{AE}{AD}=\dfrac23$

$\to\dfrac{AE}{AD-AE}=\dfrac2{3-2}$

$\to \dfrac{AE}{ED}=\dfrac{2}{1}$

Mà $CD//AB\to \dfrac{AB}{DF}=\dfrac{EA}{ED}=\dfrac21$

$\to DF=\dfrac12AB=6$

$\to S_{ADF}=\dfrac12AD\cdot AF=36$

d.Kẻ $BH\perp BE, H\in CD$

Ta có $\widehat{EBI}=45^o\to BI$ là phân giác $\widehat{EBH}$

Xét $\Delta BAE,\Delta BCH$ có;

$\widehat{BAE}=\widehat{BCH}=90^o$

$BA=BC$

$\widehat{ABE}=90^o-\widehat{EBC}=\widehat{CBH}$

$\to\Delta BAE=\Delta BCH(g.c.g)$

$\to BE=BH, CH=AE=8\to DE=4$

Xét $\Delta BEI, \Delta BHI$ có:

Chung $BI$

$\widehat{EBI}=\widehat{HBI}$ vì $BI$ là phân giác $\widehat{EBH}$

$BE=BH$

$\to\Delta BEI=\Delta BHI(c.g.c)$

$\to S_{BEI}=S_{BHI}=\dfrac12\cdot BC\cdot HI$

Mặt khác $EI=IH=IC+CH=(CD-ID)+8=(12-ID)+8=20-ID$

Ta có $\Delta DEI$ vuông tại $D$

$\to EI^2=ED^2+DI^2$

$\to (20-ID)^2=4^2+DI^2$

$\to ID=\dfrac{48}{5}$

$\to IH=20-ID=20-\dfrac{48}{5}=\dfrac{52}{5}$

$\to S_{BIH}=\dfrac12\cdot 12\cdot \dfrac{52}{5}=\dfrac{312}{5}$

$\to S_{BEI}=\dfrac{312}{5}$

e.Từ câu $d\to BE=BH$

Ta có $BE\perp BH\to\Delta BHF$ vuông tại $B$

Mà $BC\perp CD\to BC\perp FH$

$\to BC\cdot HF=BH\cdot BF(=2S_{BHF})$

$\to \dfrac{HF^2}{HF^2\cdot BC^2}=\dfrac{1}{BC^2}$

$\to \dfrac{HF^2}{(HF\cdot BC)^2}=\dfrac{1}{BC^2}$

$\to \dfrac{BF^2+BE^2}{(BH\cdot BF)^2}=\dfrac{1}{BC^2}$

$\to \dfrac{BF^2+BE^2}{BH^2\cdot BF^2}=\dfrac{1}{BC^2}$

$\to \dfrac{1}{BH^2}+\dfrac{1}{BF^2}=\dfrac{1}{BC^2}$

$\to \dfrac{1}{BE^2}+\dfrac{1}{BF^2}=\dfrac{1}{BC^2}$ không đổi