Answer:
Here we proved BE = DE and AE = CE.
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