Đáp án:

a) $\triangle AHB\backsim\triangle CAB, AH^2=AB.AD$

b) $AB.AD=AE.AC$

c) $\dfrac{S_{\triangle ADE}}{S_{\triangle ACB}}=\dfrac{144}{625}$

Giải thích các bước giải:

a)

Xét $\triangle AHB$ và $\triangle CAB$:

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

$\widehat{B}$: chung

$\to\triangle AHB\backsim\triangle CAB$ (g.g)

Xét $\triangle ADH$ và $\triangle AHB$:

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

$\widehat{DAH}$: chung

$\to\triangle ADH\backsim\triangle AHB$ (g.g)

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

b)

Xét $\triangle AEH$ và $\triangle AHC$:

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

$\widehat{EAH}$: chung

$\to\triangle AEH\backsim\triangle AHC$ (g.g)

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

Mà $AH^2=AB.AD$ (cmt)

$\to AB.AD=AE.AC$

c)

Ta có: $AB.AD=AE.AC$

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

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

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

$\widehat{EAD}$: chung

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

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

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

$\triangle AHB\backsim\triangle CAB$ (cmt)

$\to\dfrac{AH}{AB}=\dfrac{CA}{CB}\\\to AH=\dfrac{AB.CA}{CB}=\dfrac{6.8}{10}=4,8(cm)$

$AH^2=AB.AD\\\to AD=\dfrac{AH^2}{AB}=\dfrac{4,8^2}{6}=3,84(cm)$

$\triangle ADE\backsim\triangle ACB$ (cmt)

$\to k=\dfrac{AD}{AC}=\dfrac{3,84}{8}=\dfrac{12}{25}\\\to\dfrac{S_{\triangle ADE}}{S_{\triangle ACB}}=k^2=\dfrac{144}{625}$