a.
\(tan\widehat{ACD}=\dfrac{AD}{CD}=\dfrac{1}{2}\Rightarrow\widehat{ACD}\approx26^034'\)
b.
Pitago tam giác vuông ACD:
\(AD^2+CD^2=AC^2\Leftrightarrow AD^2+\left(2AD\right)^2=20\)
\(\Leftrightarrow AD^2=4\Rightarrow AD=2\)
\(AB=AD=2\) ; \(CD=2AB=4\)
c.
Hệ thức lượng:
\(DH.AC=AD.CD\Rightarrow DH=\dfrac{AD.CD}{AC}=\dfrac{4\sqrt{5}}{5}\)
Hệ thức lượng:
\(AD^2=AH.AC\Rightarrow AH=\dfrac{AD^2}{AC}=\dfrac{2\sqrt{5}}{5}\)
\(\Rightarrow\dfrac{AH}{AB}=\dfrac{\sqrt{5}}{5}\) ; \(\dfrac{AB}{AC}=\dfrac{\sqrt{5}}{5}\) \(\Rightarrow\dfrac{AH}{AB}=\dfrac{AB}{AC}\)
Xét hai tam giác ABH và ACB có:
\(\left\{{}\begin{matrix}\dfrac{AH}{AB}=\dfrac{AB}{AC}\left(cmt\right)\\\widehat{HAB}\text{ chung}\end{matrix}\right.\) \(\Rightarrow\Delta ABH\sim\Delta ACB\) (c.g.c)
\(\Rightarrow\widehat{ABH}=\widehat{ACB}\)
