SOLUTION
Consider the image in the below
From the diagram above,
[tex]\begin{gathered} H\text{ is the mid point of AD} \\ \text{And } \\ G\text{ is the midpoint of AE} \end{gathered}[/tex]
Then we obtain the coordinate of H and G
Using the coordinate of midpoint formula,
[tex]\begin{gathered} \text{Coordinate of H is } \\ (-\frac{2h+0}{2},\frac{2k+0}{2})=(\frac{-2h}{2},\frac{2k}{2})=(-h,k) \end{gathered}[/tex]
Then
[tex]\begin{gathered} \text{Coordinate of G } \\ (\frac{2h+0}{2},\frac{2k+0}{2})=(\frac{2h}{2},\frac{2k}{2})=(h,k) \end{gathered}[/tex]
Then use the distance formula to fined the lenght of |EH| and |DG|
The Distance formula is given by
[tex]\text{Distance}=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
Hence
[tex]\begin{gathered} U\sin g\text{ the coordinatesH= (-h,k) and E= (2h,0) } \\ \text{Then} \\ |EH|=\sqrt[]{(2h-(-h)^2+(0-k)^2} \\ |EH|=\sqrt[]{(3h)^2+(-k)^2} \\ \text{Then } \\ |EH|=\sqrt[]{9h^2+k^2} \end{gathered}[/tex]
Then
[tex]\begin{gathered} \text{ Using the coordinatesG= (h,k) and D=(-2h,0) for |DG|} \\ |DG|\text{ =}\sqrt[]{(-2h-h)^2+(0-k)^2} \\ |DG|=\sqrt[]{(-3h)^2^{}+(-k)^2} \\ \text{hence } \\ |DG|-=\sqrt[]{9h^2+k^2} \end{gathered}[/tex]
Hence
Since we obtain the same expression above
Then
[tex]\begin{gathered} |EH|=|DG| \\ or \\ |DG|=|EH| \end{gathered}[/tex]
therefore
[tex]|DG|\cong|EH|[/tex]
Therefore
Answer: DG is congruent to side EH