a) Xét `ΔAOB` VÀ `ΔDAB` có:

$\widehat{AOB}$`=`$\widehat{DAB}$`=`$90^o$

$\widehat{ABD}$ chung

`=>``ΔOAB`$\backsim$`ΔDAB` `(g.g)`

b) Vì `ΔAOB`$\backsim$`ΔDAB` `(g.g)`

`=>`$\widehat{OAB}$`=`$\widehat{ADB}$         `(1)`

Vì `AB``CD` ( Do `ABCD` là hình thang vuông )

`=>` $\widehat{BAC}$`=`$\widehat{ACD}$ ( 2 góc so le trong )         `(2)`

Từ `(1)` và `(2)` `=>` $\widehat{ADB}$`=`$\widehat{ACD}$

Xét `ΔBAD` và `ΔADC` có:

$\widehat{ADB}$`=`$\widehat{ACD}$ (cmt)

$\widehat{BAD}$`=`$\widehat{ADC}$`=`$90^o$

`=>``ΔBAD`$\backsim$`ΔADC` `(g.g)`

`=>` $\dfrac{AD}{DC}$`=`$\dfrac{AB}{AD}$ 

`=>` $AD^{2}$`=``AB.DC``=`$AD^{2}$`=``4.9``=`$AD^{2}$`=``36`

`=>``AD=6` `(cm)`

c) Xét `ΔOAB` và `ΔOCD` có:

$\widehat{AOB}$`=`$\widehat{COD}$`=`$90^o$

$\widehat{BAC}$`=`$\widehat{ACD}$ (cmt)

`=>``ΔOAB`$\backsim$`ΔOCD` `(g.g)`

`=>`$\dfrac{OA}{OC}$`=`$\dfrac{OB}{OD}$

`=>``OA.OD=OB.OC`

d) Vì `AB``CD`

`=>`$\dfrac{AB}{CD}$`=`$\dfrac{OA}{OC}$

`=>` $\dfrac{OA}{OC}$`=`$\dfrac{4}{9}$ 

Vì `Δ OAB`$\backsim$`ΔOCD` `(g.g)`

`=>` $\dfrac{SΔ_{OAB} }{SΔ_{OCD}}$`=`$k^{2}$

`=>` $\dfrac{SΔ_{OAB} }{SΔ_{OCD}}$`=`$\dfrac{4^{2} }{9^{2} }$`=`$\dfrac{16}{81}$ 

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