Đáp án:

a) $\triangle CED\backsim\triangle CAB$

b) $\dfrac{CD}{DE}=\dfrac{5}{3}$

c) $S_{\triangle ABD}=\dfrac{162}{7}(cm^2)$

Giải thích các bước giải:

a)

Xét $\triangle CED$ và $\triangle CAB$:

$\widehat{CED}=\widehat{CAB}\,\,\,(=90^o)$

$\widehat{C}$: chung

$\to\triangle CED\backsim\triangle CAB$ (g.g)

b)

$\triangle ABC$ vuông tại A:

$AB^2+AC^2=BC^2$ (định lý Pytago)

$\to BC=\sqrt{AB^2+AC^2}=\sqrt{9^2+12^2}=15(cm)$

$\triangle CED\backsim\triangle CAB$ (cmt)

$\to\dfrac{CD}{DE}=\dfrac{CB}{BA}=\dfrac{15}{9}=\dfrac{5}{3}$

c)

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

$\to\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{9}{12}=\dfrac{3}{4}\\\to DB=\dfrac{3}{4}DC$

Ta có:

$DB+DC=BC=15(cm)\\\to\dfrac{3}{4}DC+DC=15\\\to 7DC=60\\\to DC=\dfrac{60}{7}(cm)$

Lại có:

$\dfrac{CD}{DE}=\dfrac{5}{3}\\\to DE=\dfrac{3CD}{5}=\dfrac{3.\dfrac{60}{7}}{5}=\dfrac{36}{7}(cm)$

$\to S_{\triangle ADC}=\dfrac{1}{2}.DE.AC=\dfrac{1}{2}.\dfrac{36}{7}.12=\dfrac{216}{7}(cm^2)$

$S_{\triangle ABC}=\dfrac{1}{2}.AB.AC=\dfrac{1}{2}.9.12=54(cm^2)\\\to S_{\triangle ABD}=54-\dfrac{216}{7}=\dfrac{162}{7}(cm^2)$