Given: AE ≅ CE ; DE ≅ BE Prove: ABCD is a parallelogram. Parallelogram A B C D is shown. Diagonals are drawn from point A to poi
nt C and from point D to point B and intersect at point E. Line segments A E and E C are congruent. Line segments D E and E B are congruent. We have that AB || DC. By a similar argument used to prove that △AEB ≅ △CED, we can show that △ ≅ △CEB by. So, ∠CAD ≅ ∠ by CPCTC. Therefore, AD || BC by the converse of the theorem. Since both pair of opposite sides are parallel, quadrilateral ABCD is a parallelogram.
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 a parallelogram bisect each other.
In ΔACD and ΔBEC
AD=BC (∵Opposite sides of a 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
