Giải thích các bước giải:
a.Ta có $AC=\sqrt{AB^2+BC^2}=13$
$\sin\widehat{BAC}=\dfrac{BC}{AC}=\dfrac{12}{13}\to \widehat{BAC}=\arcsin\dfrac{12}{13}\to \widehat{ACB}=90^o-\widehat{BAC}$
b.$BH.AC=AB.BC\to BH=\dfrac{60}{13}$
$AB^2=AH.AC\to AH=\dfrac{25}{13}$
$HC=AC-HA=\dfrac{144}{13}$
c.Ta có $CH\perp BI,BC\perp CI\to BC^2=BH.BI$
Mà $BH\perp AC\to BC^2=CH.CA\to BH.BI=CH.CA$
d.Ta có :
$\dfrac{HN}{BH}=\dfrac{AH}{HC}=\dfrac{BH}{HI}\to BH^2=HN.HI$ (AB//CD, AD//BC)