Consider the paragraph proof. given: d is the midpoint of ab, and e is the midpoint of ac. prove:de = one-halfbc on a coordinate plane, triangle a b c is cut by line segment d e. point d is the midpoint of side a b and point e is the midpoint of side a c. point a is at (2 b, 2 c), point e is at (a b, c), point c is at (2 a, 0), point b is at (0, 0), and point d is at (b, c). it is given that d is the midpoint of ab and e is the midpoint of ac. to prove that de is half the length of bc, the distance formula, d = startroot (x 2 minus x 1) squared (y 2 minus y 1) squared endroot, can be used to determine the lengths of the two segments. the length of bc can be determined with the equation bc = startroot (2 a minus 0) squared (0 minus 0) squared endroot, which simplifies to 2a. the length of de can be determined with the equation de = startroot (a b minus b) squared (c minus c) squared endroot, which simplifies to ________. therefore, bc is twice de, and de is half bc. which is the missing information in the proof? a 4a a2 4a2

Given d is the mid point of ab and e is the mid point of ac. Coordinates are point a (2b,2c), point e (ab, c), point c (2a, 0),point b (0,0), point d (b, c).

We have to find the missing proof in the solution.

To find the missing figure we have to just find the distance between point d and point e.

Distance formula for determining the distance between two points on a coordinate plane is given as :

d=[tex]\sqrt{(y_{2} -y_{1} )^{2} +(x_{2} -x_{1} )^{2} }[/tex]

where ([tex]x_{1} ,y_{1} ) (x_{2} ,y_{2} )[/tex] are the coordinates at the end of line.

DE=[tex]\sqrt{(a+b-b)^{2} +(c-c)^{2} }[/tex]

Simplifying this we get:

DE=[tex]\sqrt{(a^{2} +0^{2} }[/tex]

=[tex]\sqrt{a^{2} }[/tex]

=a

Hence the missing proof is a which is the distance or length of DE..

Learn more about distance at https://brainly.com/question/2854969

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