Đáp án:

a) $\triangle AHD\backsim\triangle ABH$

b) $AD.AB=AC.AE$

c) $\dfrac{S_{\triangle ADE}}{S_{\triangle ACB}}=\dfrac{4}{25}$

Giải thích các bước giải:

a)

Xét $\triangle AHD$ và $\triangle ABH$:

$\widehat{ADH}=\widehat{AHB}\,\,\,(=90^o)$

$\widehat{HAB}$: chung

$\to\triangle AHD\backsim\triangle ABH$ (g.g)

$\to\dfrac{AH}{AD}=\dfrac{AB}{AH}\\\to AD.AB=AH^2$

b)

Xét $\triangle AHE$ và $\triangle ACH$:

$\widehat{AEH}=\widehat{AHC}\,\,\,(=90^o)$

$\widehat{HAC}$: chung

$\to\triangle AHE\backsim\triangle ACH$ (g.g)

$\to\dfrac{AH}{AE}=\dfrac{AC}{AH}\\\to AC.AE=AH^2$

Mà $AD.AB=AH^2$ (cmt)

$\to AD.AB=AC.AE$

c)

$AD.AB=AC.AE$ (cmt)

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

Xét $\triangle ADE$ và $\triangle ACB$:

$\dfrac{AD}{AC}=\dfrac{AE}{AB}$ (cmt)

$\widehat{BAC}$: chung

$\to\triangle ADE\backsim\triangle ACB$ (c.g.c)

$\to k=\dfrac{AE}{AB}$

$\triangle AHB$ vuông tại H:

$AH^2+HB^2=AB^2$ (định lý Pytago)

$\to AB=\sqrt{AB^2+HB^2}=\sqrt{6^2+8^2}=10(cm)$

$\to k=\dfrac{AE}{AB}=\dfrac{4}{10}=\dfrac{2}{5}\\\to \dfrac{S_{\triangle ADE}}{S_{\triangle ACB}}=k^2=\dfrac{4}{25}$