Question

Given: AD ≅ BC, start overline, AB ≅ CD and CE≅ BE. Prove: △FBE≅△DCE.

Answer 1

I assume the missing intended givens are that AB ∥ CD and AD ∥ BC (so ABCD is a parallelogram) and that F is the midpoint of AD (AF ≅ FD) while E is the midpoint of BC (CE ≅ BE). With those, prove △FBE ≅ △DCE.

Proof (vector/coordinate style, short and clear):

Put A at the origin and let vectors u = AB and v = AD. Then
- B = A + u = u,
- D = A + v = v,
- C = A + u + v = u + v.
Midpoints:
- F, midpoint of AD, is F = A + v/2 = v/2,
- E, midpoint of BC, is E = B + (C − B)/2 = u + v/2.

Compute the vectors of the three sides of triangle FBE and the corresponding three sides of triangle DCE:
- FE = E − F = (u + v/2) − v/2 = u,
- DC = C − D = (u + v) − v = u, so FE = DC.
- BF = F − B = v/2 − u = −(u − v/2),
- DE = E − D = (u + v/2) − v = u − v/2, so |BF| = |DE|.
- BE = E − B = (u + v/2) − u = v/2,
- CE = C − E = (u + v) − (u + v/2) = v/2, so BE = CE.

Thus FE = DC, BF = DE and BE = CE. By SSS, △FBE ≅ △DCE. ∎

(Geometric wording: in the parallelogram ABCD the segment joining the midpoints of AD and BC equals and is parallel to AB (hence to DC), and the other corresponding segments match as above; therefore the two triangles are congruent by SSS.)