Proving the Parallelogram Diagonal Theorem

Given ABCD is a parralelogam, Diagnals AC and BD intersect at E

Prove AE is conruent to CE and BE is congruent to DE

Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of parallelogram bisect each other.

In ΔACD and ΔBEC

AD=BC (∵Opposite sides of parallelogram are equal)

∠DAC=∠BCE (∵Alternate angles)

∠ADC=∠CBE (∵Alternate angles)

By ASA rule, ΔACD≅ΔBEC

By CPCT(Corresponding Parts of Congruent triangles)

AE=EC and DE=EB

Hence, AE is conruent to CE and BE is congruent to DE

olayemiolakunle65 olayemiolakunle65 · Answer 2 · 2021-10-06T21:30:54+02:00

A parallelogram is an example of quadrilaterals that have opposite sides to be equal. Its two diagonals bisect each other.

The required prove is stated below:

The diagonals of a parallelogram are bisectors of each other. This implies that the diagonals divides each other into two equal parts.

Comparing ΔABE and ΔDCE;

AB = DC (given sides of the parallelogram)

<ABC = <BDC (alternate angle property)

<BAC = DCA (alternate angle property)

Then,

AC = AE + EC

So that,

AE = EC = [tex]\frac{AC}{2}[/tex] (diagonal property of a parallelogram)

Also,

DB = DE + BE

So that,

DE = BE = [tex]\frac{DB}{2}[/tex] (diagonal property of a parallelogram)

Therefore in the given diagonal:

AE = CE and CE = BE

A sketch of the parallelogram is attached to this answer for more clarifications.

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Proving the Parallelogram Diagonal TheoremGiven ABCD is a parralelogam, Diagnals AC and BD intersect at EProve AE is conruent to CE and BE is congruent to DE

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Proving the Parallelogram Diagonal Theorem

Given ABCD is a parralelogam, Diagnals AC and BD intersect at E

Prove AE is conruent to CE and BE is congruent to DE