Given abcd is a parallelagram diagonals ac and bd intersect at e prove ae is congruent to ce and be is congruent to de

In the parallelogram ABCD, diagonals AC and BD intersect at point E.

According to the definition of parallelogram, opposite sides are equal and parallel to each other. That means, AB = DC

Now as AB and DC are parallel, so according to the property of Alternate Interior Angles, we will get:

∠EAB = ∠ECD and ∠EBA = ∠EDC

Thus , in two triangles ΔABE and ΔDCE, two angles and one side are equal. So, ΔABE and ΔDCE are congruent to each other.

That means, AE = CE and BE = DE

So, AE is congruent to CE and BE is congruent to DE

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