Complete the coordinate proof of the theorem.

Given: A B C D is a parallelogram. Prove: The diagonals of A B C D bisect each other.

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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.

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PandasSurvive PandasSurvive · Answer 1 · 2017-11-08T20:15:10+01:00

did you try ac is a+c/2,b/2? and bd is a+c/2,b/2 because i think that may be the answer

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jbiain jbiain · Answer 2 · 2020-04-09T02:53:06+02:00

Answer:

The coordinates of parallelogram ABCD are A(0, 0) , B(a, 0) , C(a+c, b), and D(c, b)

The coordinates of the midpoint of AC are ((a+c)/2, b/2)

The coordinates of the midpoint of BD are ((a+c)/2, b/2)

Step-by-step explanation:

The x-coordinate of C is the addition of x-coordinates of D and B

The y-coordinate of C is the same as y-coordinates of D

The x-coordinate of the midpoint of AC are half of the distance in x-coordinate between A and C, that is, (a+c)/2

The y-coordinate of the midpoint of DB are half of the distance in y-coordinate between D and B, that is, b/2