Write a coordinate proof: Given: Coordinates of triangle DEF, H is the midpoint of DA, G is the midpoint of EA . Prove: Side DG is congruent to side EH Here is the image down below. I have to fill in the blanks with each of the correct choices that are provided down below.

Write a coordinate proof Given Coordinates of triangle DEF H is the midpoint of DA G is the midpoint of EA Prove Side DG is congruent to side EH Here is the ima class=

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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



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