Asked by Ian
Given: AD ≅ BC, AB ≅ CD and CE ≅ BE. Prove: △FBE≅△DCE.
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(Assume F is the midpoint of AD and E is the midpoint of BC — the statement CE ≅ BE says E is the midpoint of BC.)
1. From AD ≅ BC and AB ≅ CD we have both pairs of opposite sides equal, so ABCD is a parallelogram. Put A at the origin and use position vectors: let B = b and D = d, so C = B + D = b + d.
2. Midpoints: F = (A + D)/2 = d/2, and E = (B + C)/2 = (b + (b + d))/2 = b + d/2.
3. Compute the three side vectors of the two triangles:
- FE = E − F = (b + d/2) − d/2 = b.
- DC = C − D = (b + d) − d = b.
Hence FE = DC.
- FB = B − F = b − d/2.
- DE = E − D = (b + d/2) − d = b − d/2.
Hence FB = DE.
- By hypothesis BE = CE.
4. Thus the corresponding three sides of △FBE and △DCE are pairwise equal (FB = DE, BE = CE, FE = DC). Therefore △FBE ≅ △DCE by SSS.
QED.
1. From AD ≅ BC and AB ≅ CD we have both pairs of opposite sides equal, so ABCD is a parallelogram. Put A at the origin and use position vectors: let B = b and D = d, so C = B + D = b + d.
2. Midpoints: F = (A + D)/2 = d/2, and E = (B + C)/2 = (b + (b + d))/2 = b + d/2.
3. Compute the three side vectors of the two triangles:
- FE = E − F = (b + d/2) − d/2 = b.
- DC = C − D = (b + d) − d = b.
Hence FE = DC.
- FB = B − F = b − d/2.
- DE = E − D = (b + d/2) − d = b − d/2.
Hence FB = DE.
- By hypothesis BE = CE.
4. Thus the corresponding three sides of △FBE and △DCE are pairwise equal (FB = DE, BE = CE, FE = DC). Therefore △FBE ≅ △DCE by SSS.
QED.
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