Giải thích các bước giải:
a.i)Ta có $\Delta ABD$ vuông tại $A$
$\to BD=\sqrt{AB^2+AD^2}=15$
Mà $\sin\widehat{ABD}=\dfrac{AD}{BD}=\dfrac35$
$\to\widehat{ABD}\approx 36.7^o$
$\to\widehat{ADB}=90^o-\widehat{ABD}\approx 53.3^o$
ii)Ta có $AC\perp BD\to AO\perp DB$
Mà $\Delta ABD$ vuông tại $A$
$\to AB^2=BO\cdot BD$
$\to BO=\dfrac{AB^2}{BD}=\dfrac{27}5\to DO=BD-BO=\dfrac{48}5$
$\to AO=\sqrt{AB^2-BO^2}=\dfrac{36}5$
Mà $AD\perp DC, DO\perp AC$
$\to AD^2=AO\cdot AC$
$\to AC=\dfrac{AD^2}{AO}=20$
iii)Ta có: $BH\perp CD, AB\perp AD, AD\perp DC$
$\to ABHD$ là hình chữ nhật
$\to BH=AD=12, DH=AB=9$
Ta có: $OC=AC-AO=\dfrac{64}5$
$\to CD=\sqrt{OD^2+OC^2}=16$
Ta có: $S_{OCD}=\dfrac12OD\cdot OC=\dfrac12OD\cdot (AC-AO)=\dfrac{1536}{25}$
$\to S_{DOH}=\dfrac{DH}{DC}\cdot S_{OCD}=\dfrac{864}{25}$
b.Xét $\Delta ABD,\Delta ACD$ có:
$\widehat{BAD}=\widehat{ADC}(=90^o)$
$\widehat{ADB}=90^o-\widehat{ODC}=\widehat{OCD}=\widehat{ACD}$
$\to\Delta ABD\sim\Delta DAC(g.g)$
$\to\dfrac{AD}{DC}=\dfrac{AB}{DA}$
$\to AD^2=AB\cdot CD$
Mà $BH=AD\to BH^2=AB\cdot CD$
