Đáp án:
a) AB = AC= AD=10
góc B = 60 độ
=> tam giác ABC đều
=> BC=10cm
Tam giác ABD vuông cân tại A có:
$\begin{array}{l}
B{D^2} = A{B^2} + A{D^2}\\
\Rightarrow B{D^2} = {10^2} + {10^2} = 200\\
\Rightarrow BD = 10\sqrt 2 \left( {cm} \right)\\
b)\Delta BHA \bot A;\widehat {HAB} = {60^0}\\
\sin \widehat {HAB} = \dfrac{{BH}}{{AB}}\\
\Rightarrow \sin {60^0} = \dfrac{{BH}}{{10}}\\
\Rightarrow BH = 10.\dfrac{{\sqrt 3 }}{2} = 5\sqrt 3 \left( {cm} \right)\\
Do:\widehat {HAB} + \widehat {DAK} = \widehat {BAD} = {90^0}\\
\Rightarrow \widehat {DAK} = {30^0}\\
Trong:\Delta DAK \bot K;\widehat {DAK} = {30^0}\\
\Rightarrow DK = \dfrac{{DA}}{2} = 5\left( {cm} \right)\\
c)HA = \dfrac{{BA}}{2} = 5\left( {cm} \right)\\
AK = \sqrt {A{D^2} - D{K^2}} = 5\sqrt 3 \left( {cm} \right)\\
\Rightarrow HK = AK - HA = 5\sqrt 3 - 5\left( {cm} \right)\\
d)Trong:\Delta BEC \bot E\\
\widehat {ECB} = {180^0} - \widehat {BCA} - \widehat {ACD} = {60^0}\\
\Rightarrow \sin \widehat {ECB} = \dfrac{{BE}}{{BC}}\\
\Rightarrow BE = \sin {60^0}.BC = \dfrac{{\sqrt 3 }}{2}.10 = 5\sqrt 3 \left( {cm} \right)\\
\Rightarrow CE = 5\left( {cm} \right)\\
+ Trong:\Delta BED \bot E\\
\Rightarrow DE = \sqrt {B{D^2} - B{E^2}} \\
= \sqrt {{{\left( {10\sqrt 2 } \right)}^2} - {{\left( {5\sqrt 3 } \right)}^2}} = 5\sqrt 5 \left( {cm} \right)\\
\Rightarrow DC = DE - CE = 5\sqrt 5 - 5\left( {cm} \right)
\end{array}$
