Question

Answer 1

Cho $AB=12,\ AC=18,\ BC=27,\ CD=12$
=> $BD=BC-CD=27-12=15$

a) Ta có:
$\dfrac{AC}{CB}=\dfrac{18}{27}=\dfrac{2}{3}$

$\dfrac{DC}{CA}=\dfrac{12}{18}=\dfrac{2}{3}$

Lại có $\widehat{ACB}=\widehat{DCA}$ (góc chung)

$\Rightarrow \triangle ACB \sim \triangle DCA$ (theo c.g.c)

b) Từ $\triangle ACB \sim \triangle DCA$ suy ra:
$\dfrac{AB}{AD}=\dfrac{AC}{DC}$

$\Rightarrow \dfrac{12}{AD}=\dfrac{18}{12}=\dfrac{3}{2}$

$\Rightarrow AD=\dfrac{12 \cdot 2}{3}=8\text{ cm}$

c) Do $\triangle ACB \sim \triangle DCA$ nên các góc tương ứng bằng nhau, suy ra:
$\widehat{DAC}=\widehat{ABC}=70^\circ$