Given: ABCD is a parallelogram.

Diagonals AC, BD intersect at E.

Prove: AE = CE and BE = DE

Here we proved BE = DE and AE = CE.

Step-by-step explanation:

Given that,

Diagram of the given scenario is below.

ABCD is a parallelogram.

Diagonals AC, BD intersect at E.

We have to prove AC = CE and BE = DE.

Now,

Taking ΔDEC and ΔAEB,

DC = AB                     [opposite sides of parallelogram are equal]

∠CDE = ∠ABE           [alternate interior angles are equal]

∠DEC = ∠AEB           [vertically opposite angles are equal]

∴ ΔDEC ≅ ΔAEB.

By CPCT                   [corresponding part of congruent triangle]

DE = BE                  ...(i)

Taking ΔAED and ΔBEC,

AD = BC                    [opposite sides of parallelogram are equal]

∠AED = ∠BEC          [vertically opposite angles are equal]

∠EAD = ∠ECB          [alternate interior angles are equal]

∴ ΔAED ≅ ΔBEC.

By CPCT                   [corresponding part of congruent triangle]

AE = CE            ...(ii)

From equation (i) and (ii),

BE = DE and AE = CE

I don't like these very much

Step-by-step explanation:

Sorry

38.6°

step-by-step explanation:

resultant is 10 pound

smaller force = 8 pound

angle formed = tan⁻¹(

                        =tan⁻¹(

                          =tan⁻¹(0.8)

                        =38.6°