Giải thích các bước giải:
a.Ta có : $AH\perp BC\to\widehat{AHB}=\widehat{BAC}=90^o$
$\to \Delta AHB\sim\Delta CAB(g.g)$
b.Tương tự câu a $\to \Delta AHC\sim\Delta BAC(g.g)$
$\to\dfrac{AC}{BC}=\dfrac{HC}{AC}\to AC^2=CH.BC$
c.Vì $\widehat{AHD}=45^o, \widehat{AHC}=90^o\to HD$ là phân giác $\widehat{AHC}$
$\to\dfrac{DA}{DC}=\dfrac{HA}{HC}$
Từ câu a,b $\to \widehat{BAH}=\widehat{ACH},\widehat{CAH}=\widehat{ABH}$
$\to\Delta AHB\sim\Delta CHA(g.g)$
$\to\dfrac{HA}{HC}=\dfrac{AB}{AC}$
$\to \dfrac{DA}{DC}=\dfrac{AB}{AC}$
d.Từ câu c
$\to \dfrac{P_{AHB}}{P_{CHA}}=\dfrac{AH}{CH}=\dfrac{HB}{HA}=\dfrac{AB}{AC}=\dfrac{15}{20}=\dfrac34$
$\to CH=\dfrac43AH, BH=\dfrac34AH$
$\to AB=\sqrt{AH^2+HB^2}=\dfrac54AH$
Mà $\Delta AHB\sim\Delta CAB$
$\to \dfrac{P_{ABC}}{P_{AHB}}=\dfrac{AB}{HB}=\dfrac{\dfrac54AH}{\dfrac34AH}=\dfrac53$
$\to P_{ABC}=\dfrac53P_{AHB}=25$
