Giải thích các bước giải:
a.Ta có $\Delta ABC$ vuông tại $A$
$\to BC=\sqrt{AB^2+AC^2}=5$
Mà $HA\perp BC$
$\to AB^2=BH\cdot BC$
$\to BH=\dfrac{AB^2}{BC}=\dfrac{9}5$
$\to CH=BC-BH=\dfrac{16}5$
b.Ta có $AB\perp AC, CE\perp BE, CE\perp AC$
$\to ABEC$ là hình chữ nhật
$\to BE=AC=4, CE=AB=3$
Ta có $CD\perp AC, CH\perp AD$
$\to AC^2=AH\cdot AD$
$\to AD=\dfrac{AC^2}{AH}=\dfrac{20}3$
$\to CD=\sqrt{AD^2-AC^2}=\dfrac{16}3$
$\to S_{BCD}=\dfrac12EB\cdot CD=\dfrac{32}3$
c.Xét $\Delta ABC,\Delta ACD$ có:
$\widehat{BAC}=\widehat{ACD}(=90^o)$
$\widehat{ACB}=90^o-\widehat{BCD}=\widehat{HDC}=\widehat{ADC}$
$\to\Delta ABC\sim\Delta CAD(g.g)$
$\to\dfrac{AC}{CD}=\dfrac{AB}{CA}$
$\to AC^2=AB\cdot CD$
d.Ta có $HI\perp AC\to HI//AB//CD$ vì $AB\perp AC, AC\perp CD$
$\to \dfrac{HI}{AB}=\dfrac{CH}{CB}$
$\to HI=\dfrac{CH\cdot AB}{CB}$
Lại có $HK//CD$
$\to \dfrac{HK}{CD}=\dfrac{BH}{BC}$
$\to HK=\dfrac{BH\cdot CD}{BC}$
Do $AB//CD$
$\to \dfrac{HB}{HC}=\dfrac{AB}{CD}$
$\to CH\cdot AB=BH\cdot CD$
$\to HI=HK$
