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13+8 POINTS!!!! PLEASE HELP ME PLEASE!!!!! DUE TODAY PLEASE HELP!!!


Complete the coordinate proof of the theorem.


Enter your answers in the boxes.
The coordinates of parallelogram ABCD are A(0, 0) , B(a, 0) , C(??), and D(c, b) .

The coordinates of the midpoint of AC¯¯¯¯¯ are (, b2 ).

The coordinates of the midpoint of BD¯¯¯¯¯ are ( a+c2 , ).

The midpoints of the diagonals have the same coordinates.

Therefore, AC¯¯¯¯¯ and BD¯¯¯¯¯ bisect each other.

138 POINTS PLEASE HELP ME PLEASE DUE TODAY PLEASE HELPComplete the coordinate proof of the theoremEnter your answers in the boxes The coordinates of parallelogr class=

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Answer:

Step-by-step explanation:

1). Coordinates of point C will be

x - coordinates = [Distance of B from the origin + Distance of D from y axis]

                         = (a + c)

y - coordinates = (Distance of D from x - axis)

                         = b

Therefore, coordinates of C will be [(a + c), b]

2). Coordinates of the midpoint of AC = [tex][\frac{(a+c)+0}{2}, \frac{b+0}{2}][/tex]

= [tex][\frac{(a+c)}{2}, \frac{b}{2}][/tex]

[Since coordinates of the midpoint of (x, y) and (x', y') are denoted by ([tex]\frac{x+x'}{2},\frac{y+y'}{2}[/tex])]

3). Coordinates of the midpoint of BD =  [tex][\frac{(a+c)}{2}, \frac{b+0}{2}][/tex]

= [tex][\frac{(a+c)}{2}, \frac{b}{2}][/tex]

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