Giải thích các bước giải:

a.Ta có $\widehat{BAD}=\widehat{ADC}=90^o$  

              $\dfrac{AB}{AD}=\dfrac{4}{6}=\dfrac{6}{9}=\dfrac{AD}{CD}$
$\to\Delta BAD\sim\Delta ADC(c.g.c)$

b.Gọi $AC\cap BD=O$

Từ câu a $\to\widehat{ABD}=\widehat{DAC}$

$\to\widehat{DBA}+\widehat{BAO}=\widehat{BAO}+\widehat{CAD}=\widehat{BAD}=90^o$
$\to AC\perp BD$

c.Ta có $AB//CD\to\widehat{OAB}=\widehat{OCD},\widehat{OBA}=\widehat{ODC}$

$\to\Delta AOB\sim\Delta COD(g.g)$

$\to\dfrac{S_{AOB}}{S_{COD}}=(\dfrac{AB}{CD})^2=(\dfrac49)^2=\dfrac{16}{81}$

d.Ta có : $AB//CD$

$\to\dfrac{KA}{KD}=\dfrac{AB}{CD}$

$\to\dfrac{KA}{KD-KA}=\dfrac{AB}{CD-AB}$

$\to\dfrac{KA}{AD}=\dfrac{AB}{CD-AB}$

$\to KA=\dfrac{AB.AD}{CD-AB}$

$\to KA=\dfrac{4.6}{9-4}=\dfrac{24}{5}$