hình học lớp 8

P

phamhuy20011801

$a, \dfrac{9}{6+7,5}=\dfrac{9}{13,5}=\dfrac{6}{9}$
$\hat{B}$ chung.
$\rightarrow \triangle \ ABC$ ~ $\triangle \ CBD (c.g.c)$

$b, \dfrac{AC}{CD}=\dfrac{AB}{BC}=\dfrac{2}{3}$
$\rightarrow CD=\dfrac{3AC}{2}=11,25 (cm)$

c, Kẻ phân giác $AE$
$\dfrac{AB}{AC}=\dfrac{BE}{EC}=\dfrac{AB}{AD}$
$\rightarrow AE//CD \rightarrow \widehat{AEC}=\widehat{ACD}$
Hay $\widehat{ABC}=2.\widehat{ACD}$
 
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