Đáp án:

a) $\triangle BHA\backsim\triangle CHI$

b) $IC^2=IH.IB$

c) $S_{\triangle BHC}=93,75cm^2$

Giải thích các bước giải:

a)

Xét $\triangle BHA$ và $\triangle CHI$:

$\widehat{BAH}=\widehat{CIH}\,\,\,(=90^o)$

$\widehat{BHA}=\widehat{CHI}$ (đối đỉnh)

$\to\triangle BHA\backsim\triangle CHI$ (g.g)

$\to\dfrac{AH}{AB}=\dfrac{IH}{IC}$

b)

Xét $\triangle BHA$ và $\triangle BCI$:

$\widehat{BAH}=\widehat{BIC}\,\,\,(=90^o)$

$\widehat{ABH}=\widehat{IBC}$ (gt)

$\to\triangle BHA\backsim\triangle BCI$ (g.g)

$\to\dfrac{AH}{AB}=\dfrac{IC}{IB}$

Mà $\dfrac{AH}{AB}=\dfrac{IH}{IC}$ (cmt)

$\to\dfrac{IC}{IB}=\dfrac{IH}{IC}\\\to IC^2=IH.IB$

c)

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

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

$\triangle ABC$ có đường phân giác BH (gt)

$\to\dfrac{HA}{HC}=\dfrac{BA}{BC}=\dfrac{15}{25}=\dfrac{3}{5}\\\to HC=\dfrac{5}{3}HA$

Ta có:

$HA+HC=AC=20(cm)\\\to HA+\dfrac{5}{3}HA=20\\\to 8HA=60\\\to HA=7,5(cm)$

$\to S_{\triangle AHB}=\dfrac{1}{2}.AH.AB=\dfrac{1}{2}.7,5.15=56,25(cm^2)$

$S_{\triangle ABC}=\dfrac{1}{2}.15.20=150(cm^2)$

$\to S_{\triangle BHC}=S_{\triangle ABC}-S_{\triangle AHB}=150-56,25=93,75(cm^2)$