Đáp án:

a) $\triangle HAC\backsim\triangle ABC$

b) $AH^2=AD.AB$

c) $AD.AB=AE.AC$

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

Giải thích các bước giải:

a)

Xét $\triangle HAC$ và $\triangle ABC$:

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

$\widehat{C}$: chung

$\to\triangle HAC\backsim\triangle ABC$ (g.g)

b)

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

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

$\widehat{DAH}$: chung

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

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

c)

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 $AH^2=AD.AB$ (cmt)

$\to AD.AB=AE.AC$

d)

Ta có: $AD.AB=AE.AC$ (cmt)

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

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

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

$\widehat{DAE}$: chung

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

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

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

$\to BC=\sqrt{AB^2+AC^2}=\sqrt{12^2+16^2}=20(cm)$

$S_{\triangle ABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.AB.AC\\\to AH.BC=AB.AC\\\to AH=\dfrac{AB.AC}{BC}=\dfrac{12.16}{20}=9,6(cm)$

$AH^2=AD.AB$ (cmt)

$\to AD=\dfrac{AH^2}{AB}=\dfrac{9,6^2}{12}=7,68(cm)\\\to k=\dfrac{7,68}{16}=\dfrac{12}{25}$

$\to\dfrac{S_{\triangle ADE}}{S_{\triangle ACB}}=k^2=\dfrac{144}{625}$