Ta có: \(AC=AD+DC=8\left(cm\right)\)
\(\dfrac{AB}{AC}=\dfrac{4}{8}=\dfrac{1}{2}\) ; \(\dfrac{AD}{AB}=\dfrac{2}{4}=\dfrac{1}{2}\Rightarrow\dfrac{AB}{AC}=\dfrac{AD}{AB}\)
Xét 2 tam giác ADB và ABC có:
\(\left\{{}\begin{matrix}\widehat{A}\text{ chung}\\\dfrac{AB}{AC}=\dfrac{AD}{AB}\left(cmt\right)\end{matrix}\right.\) \(\Rightarrow\Delta ADB\sim\Delta ABC\left(c.g.c\right)\)
\(\Rightarrow\widehat{ABD}=\widehat{ACB}=20^0\)
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