a) \(E\) là trung điểm \(AB\) nên \(AE=EB=\dfrac{AB}{2}=18\left(cm\right)\)
Áp dụng định lí Pi-ta-go ta có:
\(DE^2=AD^2+AE^2\)
\(\Leftrightarrow DE^2=24^2+18^2\)
\(\Leftrightarrow DE=30\left(cm\right)\)
Áp dụng định lí Ta-let ta có:
\(AD\text{/ / }BC\Rightarrow AD\text{/ / }BG\Rightarrow\dfrac{DE}{EG}=\dfrac{AD}{BG}=\dfrac{AE}{EB}=1\)
\(\Rightarrow DE=EG=30\left(cm\right)\Rightarrow DG=60\left(cm\right)\)
\(AE\text{/ / }DC\Rightarrow\dfrac{AF}{FC}=\dfrac{EF}{DF}=\dfrac{AF}{DC}=\dfrac{1}{2}\)
\(\Rightarrow EF=\dfrac{1}{2}DF\Rightarrow EF=\dfrac{1}{3}DE=10\left(cm\right)\)
\(\Rightarrow DE=DE-EF=20\left(cm\right)\)
b)
Ta có :
\(FD^2=\left(\dfrac{2}{3}DE\right)^2=\dfrac{4}{9}DE^2\)
\(\Rightarrow FD^2=FE.FG\)
