Giải thích các bước giải:
1.Ta có $AH\perp BC, HM\perp AB$
$\to AM\cdot AB=AH^2$
Tương tự $AN\cdot AC=AH^2$
$\to AM\cdot AB=AN\cdot AB$
Ta có $AB\perp AC, AH\perp BC\to AH^2=HB\cdot HC$
Vì $HM\perp AB,HN\perp AC, AB\perp AC\to AMHN$ là hình chữ nhật
$\to AH=MN$
$\to MN^2=AH^2=HB\cdot HC$
2.Ta có:
$AM\cdot BM+AN\cdot CN=HM^2+HN^2=MN^2=AH^2$
3.Ta có:
$HM\cdot AB+HN\cdot AC=AH\cdot HB+AH\cdot HC=AH(HB+HC)=AH\cdot BC=AB\cdot AC$
4.Ta có:
$AH^2=HB\cdot HC$
$\to (AH^2)^2=(HB\cdot HC)^2$
$\to AH^4=HB^2\cdot HC^2$
$\to AH^4=(BM\cdot BA)\cdot (CN\cdot CA)=(BM\cdot CN)\cdot (AB\cdot AC)=BM\cdot CN\cdot (AH\cdot BC)$
$\to AH^3=BM\cdot NC\cdot BC$
5.Ta có:
$AB^2-BH^2=AH^2=AC^2-CH^2$
$\to AB^2+HC^2=AC^2+HB^2$
6.Ta có:
$\dfrac{BH}{CH}=\dfrac{BH\cdot BC}{CH\cdot BC}=\dfrac{AB^2}{AC^2}$
7.Ta có:
$MA\cdot MB+NA\cdot NC=AH^2$(câu 2)
Mà $AH^2=HB\cdot HC$
$\to MA\cdot MB+NA\cdot NC=HB\cdot HC$
8.Ta có:
$BH^3=BH^2\cdot BH=(AB^2-AH^2)\cdot BH=AB^2\cdot BH-AH^2\cdot BH=BH\cdot BC\cdot BH-AH^2\cdot BH=BH^2\cdot BC-AH^2\cdot BH$
Tương tự $CH^3=CH^2\cdot BC-AH^2\cdot CH$
Mà $MN\cdot AB\cdot AC=AH\cdot (AB\cdot CA)=AH\cdot (AH\cdot BC)=AH^2\cdot BC$
Khi đó
$ BH^3+CH^3+3MN\cdot AB\cdot AC$
$=(BH^2\cdot BC-AH^2\cdot BH)+(CH^2\cdot BC-AH^2\cdot CH)+3AH^2\cdot BC$
$=BC\cdot (BH^2+CH^2)-AH^2(HB+HC)+3AH^2\cdot BC$
$=BC\cdot (BH^2+CH^2)-AH^2BC+3AH^2\cdot BC$
$=BC\cdot (BH^2+CH^2)+2AH^2\cdot BC$
$=BC\cdot (BH^2+CH^2+2AH^2)$
$=BC\cdot (BH^2+AH^2+CH^2+AH^2)$
$=BC\cdot (AB^2+AC^2)$
$=BC\cdot BC^2$
$=BC^3$
