Question

given rectangle ABCD where E is the midpoint of BC, prove AE is = DE

Answer 1

Construction: Join AE and DE
To prove: AE = DE
Proof:

A___________D
|. |
|. |
B|.__________| C

In triangle BAE and traingle CDE,
AB=CD(opposite sides of a rectangle are equal)
BE=CE( given that E is the midpoint of BC)
AngleABE = Angle CDE ( the angles of a rectangle is 90•)
Therefore, Traingle BAE is congruent to triangle CDE by SAS congruency.
AE = DE (CPCTC)
Hence proved.

Answer 2

Answer:

Given

ABCD is rectangle and E is mid point of BC

To prove

AE = DE

proof

In ∆AEB and ∆ EDC

CE=BE ( As E is mid point of BC )

AB=DC (given )

angle AEB + angle AED = angle CED+angle AED

angle AEB =angle CED

therefore, ∆AEB is congurence to ∆ EDC [By SAS Rule ]

therefore By CPCT AE =DE