Giải thích các bước giải:

a.Ta có $CD\perp Ax\to \widehat{CDA}=\widehat{CAB}=90^o$

Mà $Ax//BC\to \widehat{DAC}=\widehat{ACB}$

$\to \Delta ADC\sim\Delta CAB(g.g)$

b. Từ câu a

$\to \dfrac{DC}{AB}=\dfrac{AC}{BC}$

$\to DC=\dfrac{AB.AC}{BC}$

Vì $\Delta ABC$ vuông tại A$\to BC^2=AB^2+AC^2=100\to BC=10$

$\to  CD=\dfrac{6.8}{10}=\dfrac{24}5$

c.Ta có $CD\perp AD\to AD^2=AC^2-CD^2=\dfrac{1024}{25}$

$\to AD=\dfrac{32}5$

Do $AD//BC\to \dfrac{IA}{IC}=\dfrac{AD}{BC}=\dfrac{16}{25}$

$\to \dfrac{IC}{IA}=\dfrac{25}{16}\to \dfrac{IC}{IC+IA}=\dfrac{25}{25+16}$

$\to \dfrac{IC}{AC}=\dfrac{25}{41}$

$\to \dfrac{S_{BIC}}{S_{ABC}}=\dfrac{25}{41}$

$\to S_{BIC}=\dfrac{25}{41}S_{ABC}=\dfrac{25}{41}\cdot \dfrac12AB\cdot AC=\dfrac{600}{41}$