Question

Answer 1

Cho $AB=8,\ AC=16,\ BD=2,\ CE=13$

Suy ra:
$AD=AB-BD=8-2=6$

$AE=AC-CE=16-13=3$

a) Xét:$\dfrac{AD}{AE}=\dfrac{6}{3}=2$

$\dfrac{AB}{AC}=\dfrac{8}{16}=2$

Lại có $\widehat{DAE}=\widehat{BAC}$ (góc chung)

$\Rightarrow \triangle AED \sim \triangle ABC$ (c.g.c)

b) Từ $\triangle AED \sim \triangle ABC$ suy ra:
$\dfrac{AD}{AC}=\dfrac{ED}{BC}$

$\Rightarrow AD \cdot BC = AC \cdot ED$

c) Do $\triangle AED \sim \triangle ABC$ nên các góc tương ứng bằng nhau, suy ra:
$\widehat{AED}=\widehat{ABC}$