Đáp án:

a) $BA^2=BH.BC, BH=9cm, CH=16cm$

b) $AM.AB=AN.AC, \triangle AMN\backsim\triangle ACB$

c) $\dfrac{S_{\triangle AMN}}{S_{\triangle ACB}}=\dfrac{144}{625}, S_{\triangle AMN}=34,56cm^2$

Giải thích các bước giải:

a)

Xét $\triangle HBA$ và $\triangle ABC$:

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

$\widehat{B}$: chung

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

$\to\dfrac{BH}{BA}=\dfrac{BA}{BC}\\\to BA^2=BH.BC$

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

$\to BC=\sqrt{AB^2+AC^2}=\sqrt{15^2+20^2}=25(cm)$

$\to BH=\dfrac{BA^2}{BC}=\dfrac{15^2}{25}=9(cm)\to CH=25-9=16(cm)$

b)

Xét $\triangle AMH$ và $\triangle AHB$:

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

$\widehat{MAH}$: chung

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

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

Chứng minh tương tự

$\to\triangle ANH\backsim\triangle AHC$

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

$\to AM.AB=AN.AC\\\to\dfrac{AM}{AC}=\dfrac{AN}{AB}$

Xét $\triangle AMN$ và $\triangle ACB$:

$\dfrac{AM}{AC}=\dfrac{AN}{AB}$ (cmt)

$\widehat{MAN}$: chung

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

c)

$\triangle HBA\backsim\triangle ABC$ (cmt)

$\to\dfrac{AH}{AB}=\dfrac{CA}{CB}\\\to AH=\dfrac{AB.CA}{CB}=\dfrac{15.20}{25}=12(cm)$

Ta có: $AH^2=AN.AC$ (cmt)

$\to AN=\dfrac{AH^2}{AC}=\dfrac{12^2}{20}=7,2(cm)$

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

$\to k=\dfrac{AN}{AB}=\dfrac{7,2}{15}=\dfrac{12}{25}\\\to\dfrac{S_{\triangle AMN}}{S_{\triangle ACB}}=k^2=\dfrac{144}{625}\\\to S_{\triangle AMN}=\dfrac{144}{625}S_{\triangle ACB}=\dfrac{144}{625}.\dfrac{1}{2}.15.20=34,56(cm^2)$