Giải thích các bước giải:

a.Xét $\Delta ABD,\Delta ACE$ có:

Chung $\hat A$

$\widehat{ADB}=\widehat{AEC}(=90^o)$

$\to \Delta ABD\sim\Delta ACE(g.g)$

b.Xét $\Delta HDC,\Delta HBE$ có:

$\widehat{EHB}=\widehat{DHC}$(đối đỉnh)

$\widehat{HEB}=\widehat{HDC}(=90^o)$

$\to \Delta HEB\sim\Delta HDC(g.g)$

$\to \dfrac{HE}{HD}=\dfrac{HB}{HC}$

$\to HB\cdot HD=HE\cdot HC$

c.Từ câu a $\to \dfrac{AB}{AC}=\dfrac{AD}{AE}$

$\to \dfrac{AD}{AB}=\dfrac{AE}{AC}$

Mà $\widehat{DAE}=\widehat{BAC}$

$\to\Delta ADE\sim\Delta ABC(c.g.c)$

$\to \widehat{ADE}=\widehat{ABC}$

d.Xét $\Delta AMD,\Delta AMC$ có:

Chung $\hat A$

$\widehat{ADM}=\widehat{AMC}(=90^o)$

$\to \Delta AMD\sim\Delta ACM(g.g)$

$\to \dfrac{AM}{AC}=\dfrac{AD}{AM}$

$\to AM^2=AD\cdot AC$

Tương tự chứng minh được $AN^2=AE\cdot AB$

Do $AD\cdot AC=AE\cdot AB \to AM^2=AN^2\to AM=AN$