in parallelogram ABCD, diagonals AC and BD intersect at E. Which statement proves ABCD is a rectangle?

AC ≅ BD or say Measure of AC = Measure of BD diagonal.

Given that:

ABCD is a parallelogram

AC and BD intersect at E.

A parallelogram is defined as a closed geometric figure that has got 4 straight sides and  opposite sides are parallel to each other.

Set of all parallelogram is superset of set of all rectangles.

There is a property about diagonals of a parallelogram that states as;

" A parallelogram is a rectangle if and only if both of its diagonals are equal in size".

Now given parallelogram is ABCD thus its diagonals are AC and BD.

Thus for ABCD to be rectangle we need:

magnitude of AC = magnitude of BC

or

AC ≅ BD

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AC ≅ BD

Step-by-step explanation:

A parallelogram is a shape that have two pairs of parallel sides with opposite sides having the same length and opposite angles having the same size.

A rectangle is a shape that have two pairs of parallel sides with opposite sides having the same length and each of the four angles is 90°.

A parallelogram is said to be a rectangle if the diagonals of the parallelogram are of the same size (congruent). Therefore in parallelogram ABCD, ABCD is a rectangle if AC is equal to BD. where AC and BD are the diagonals