Đáp án:

a) $\triangle IAC\backsim\triangle ABC$

b) $AB=5cm, IC=\dfrac{144}{13}cm, AK=\dfrac{10}{3}cm$

c) $S_{\triangle BMI}=\dfrac{625}{507}cm^2, S_{\triangle ABK}=\dfrac{25}{3}cm^2$

Giải thích các bước giải:

a)

Xét $\triangle IAC$ và $\triangle ABC$:

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

$\widehat{C}$: chung

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

b)

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

$\to AB=\sqrt{BC^2-AC^2}=\sqrt{13^2-12^2}=5(cm)$

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

$\to\dfrac{IC}{AC}=\dfrac{AC}{BC}\\\to IC=\dfrac{AC^2}{BC}=\dfrac{12^2}{13}=\dfrac{144}{13}(cm)$

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

$\to\dfrac{KC}{KA}=\dfrac{BC}{BA}=\dfrac{13}{5}\\\to KC=\dfrac{13}{5}KA$

Ta có:

$KA+KC=AC=12(cm)\\\to KA+\dfrac{13}{5}KA=12\\\to 18KA=60\\\to KA=\dfrac{10}{3}(cm)$

c)

Xét $\triangle ABK$ và $\triangle IBM$:

$\widehat{BAK}=\widehat{BIM}\,\,\,(=90^o)$

$\widehat{ABK}=\widehat{IBM}$ (gt)

$\to\triangle ABK\backsim\triangle IBM$ (g.g)

$\to\dfrac{AK}{IM}=\dfrac{AB}{IB}\\\to IM=\dfrac{AK.IB}{AB}=\dfrac{\dfrac{10}{3}.\left(13-\dfrac{144}{13}\right)}{5}=\dfrac{50}{39}(cm)$

$\to S_{\triangle BMI}=\dfrac{1}{2}.IM.IB=\dfrac{1}{2}.\dfrac{50}{39}.\left(13-\dfrac{144}{13}\right)=\dfrac{625}{507}(cm^2)\\S_{\triangle ABK}=\dfrac{1}{2}.AB.AK=\dfrac{1}{2}.5.\dfrac{10}{3}=\dfrac{25}{3}(cm^2)$