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Proving the Parallelogram Diagram Theorem
Given: ABCD is a Parallelogram. Diagonals AC,BD intersect at E.
Prove AE = CE and BE =DE
THE AWNSWERS ARE IN THE PICTURE FROM 4 DOWN SOME POSTED A PICTURE WITH THE 4 UP HERE TO HELP :)​

Proving the Parallelogram Diagram Theorem Given ABCD is a Parallelogram Diagonals ACBD intersect at E Prove AE CE and BE DE THE AWNSWERS ARE IN THE PICTURE FROM class=

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nightfurythanh12

Answer:

Did you learn about equal rectangles? we can prove AEB and CED are equal, which results in what we need to prove here

AE = CE and BE = DE. This can be proved with the help of the properties of congruent triangles.

What is a parallelogram?

'A parallelogram is a special kind of quadrilateral that is formed by parallel lines. The angle between the adjacent sides of a parallelogram may vary but the opposite sides need to be parallel for it to be a parallelogram. A quadrilateral will be a parallelogram if its opposite sides are parallel and congruent.'

According to the given problem,

ABCD is a parallelogram.

We know,

BC ║ AD and BC ≅ AD

From, the properties of a parallelogram,

∠CBD = ∠ADB

∠BCA = ∠DAC

We know, two lines are parallel and alternate interior angles of the parallelogram are equal.

Also, Δ BEC ≅ ΔAED (Angle-Side-Angle)

Therefore, AE ≅ CE ( The properties of congruent triangles )

                  BE ≅ DE ( The properties of congruent triangles )

Hence, we can conclude, in the parallelogram ABCD, AE = CE and BE = DE from the properties of congruent triangles.

Learn more about parallelogram here: https://brainly.com/question/11220936

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