a: A là trọng tâm của ΔBCD
=>\(\left\{{}\begin{matrix}x_B+x_C+x_D=3\cdot x_A\\y_B+y_C+y_D=3\cdot y_A\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1+1+x_D=3\cdot3=9\\0+4+y_D=3\cdot\left(-5\right)=-15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_D=7\\y_D=-15-4=-19\end{matrix}\right.\)
Vậy: D(7;-19)
E thuộc trục Ox nên E(x;0)
E(x;0); A(3;-5); B(1;0)
\(\overrightarrow{EA}=\left(3-x;-5\right);\overrightarrow{EB}=\left(1-x;0\right)\)
Vì ΔEAB vuông tại E nên \(\overrightarrow{EA}\cdot\overrightarrow{EB}=0\)
=>(1-x)(3-x)+(-5)*0=0
=>(1-x)(3-x)=0
=>\(\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
vậy: E(1;0); E(3;0)
c: A(3;-5); B(1;0); C(1;4)
\(AB=\sqrt{\left(1-3\right)^2+\left(0+5\right)^2}=\sqrt{29}\)
\(AC=\sqrt{\left(1-3\right)^2+\left(4+5\right)^2}=\sqrt{83}\)
\(BC=\sqrt{\left(1-1\right)^2+\left(4-0\right)^2}=4\)
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{29+83-16}{2\cdot\sqrt{29}\cdot\sqrt{83}}=\dfrac{48}{\sqrt{2407}}\)
=>\(\widehat{BAC}\simeq11^056'\)
