Since, AD ≅ BC, ∠BCD ≅ ∠ADC and segment CD ≅ DC, △ADC ≅ △BCD by SAS. Therefore, based on CPCTC, DE ≅ CE.
Explanation:
- We are given AD ≅ BC and ∠BCD ≅ ∠ADC.
- The segment CD is common for △ADC and △BCD.
- This means BCD and ADC are isosceles triangles.
- Therefore, we can say that △ADC ≅ △BCD by SAS.
- Because of CPCTC, segment DE is congruent to segment CE.
- Hence, based on the isosceles triangle theorem, DE ≅ CE.
