Giải thích các bước giải:
1.Ta có $\widehat{AMH}=\widehat{AHB}=90^o,\widehat{MAH}=\widehat{BAH}$
$\to\Delta AMH\sim\Delta AHB(g.g)$
$\to\dfrac{AM}{AH}=\dfrac{AH}{AB}$
$\to AH^2=AM\cdot AB$
Tương tự $AN\cdot AC=AH^2$
Lại có: $\widehat{AHB}=\widehat{AHC}=90^o,\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to\Delta AHB\sim\Delta CHA(g.g)$
$\to\dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to AH^2=HB\cdot HC$
$\to AM\cdot AB=AN\cdot AC=HB\cdot HC$
2.Chứng minh tương tự câu $1\to MA\cdot MB=MH^2, NA\cdot NC=NH^2$
Ta có $HM\perp AB, HN\perp AC, AB\perp AC\to AMHN$ là hình chữ nhật
$\to AH=MN$
$\t AM\cdot MB+NA\cdot NC=HM^2+HN^2=MN^2=AH^2=HB\cdot HC$
3.Ta có:
$AN\cdot AC=AH^2\to AN=\dfrac{AH^2}{AC}\to AN\cdot AB=\dfrac{AH^2\cdot AB}{AC}$
$AM\cdot AB=AH^2\to AM=\dfrac{AH^2}{AB}\to AM\cdot AC=\dfrac{AH^2\cdot AC}{AB}$
$\to AN\cdot AB+AM\cdot AC=\dfrac{AH^2\cdot AB}{AC}+\dfrac{AH^2\cdot AC}{AB}$
$\to AN\cdot AB+AM\cdot AC=AH^2\cdot (\dfrac{AB}{AC}+\dfrac{AC}{AB})$
$\to AN\cdot AB+AM\cdot AC=AH^2\cdot \dfrac{AB^2+AC^2}{AC\cdot AB}$
$\to AN\cdot AB+AM\cdot AC=AH^2\cdot \dfrac{BC^2}{AC\cdot AB}$
$\to AN\cdot AB+AM\cdot AC=\dfrac{AH^2\cdot BC^2}{AC\cdot AB}$
Vì $AH\cdot BC=AB\cdot AC=2S_{ABC}$
$\to AN\cdot AB+AM\cdot AC=\dfrac{(AB\cdot AC)^2}{AC\cdot AB}$
$\to AN\cdot AB+AM\cdot AC=AB\cdot AC$
