DE ≅ CE given that side AD and side BC are equal and angle ∠BCD and angle ∠ADC are equal. This can be obtained by using triangle congruency theorems.
Prove that side DE and side CE is equal:
Triangle congruency theorems required in the question,
- SAS triangle congruency theorem - SAS means Side-angle-side. If two sides and the included angle of a triangle is equal to two sides and the included angle of another triangle then the triangles are congruent.
- AAS triangle congruency theorem - AAS means Angle-angle-side. If two angles and one side of a triangle is equal to two angles and one side of another triangle then the triangles are congruent.
In the question it is given that,
⇒ Side DE and side CE are equal ⇒ AD ≅ BC
⇒ angle ∠BCD and angle ∠ADC are equal ⇒ ∠BCD ≅ ∠ADC
- AD ≅ BC (given in the question)
- ∠BCD ≅ ∠ADC (given in the question)
- DC ≅ DC (since DC is a common side; reflexive property)
Therefore we can say that,
ΔADC ≅ ΔBCD according to the SAS triangle congruency theorem
- ∠DAE ≅ ∠CBE (corresponding parts of congruent triangles are congruent (CPCTC))
- AD ≅ BC (given in the question)
- ∠DEA ≅ ∠CEB (since they are vertically opposite angles - vertical angles theorem)
ΔAED ≅ ΔBEC according to the AAS triangle congruency theorem
Thus DE ≅ CE since corresponding parts of congruent triangles are congruent (CPCTC).
Hence DE ≅ CE given that side AD and side BC are equal and angle ∠BCD and angle ∠ADC are equal.
Learn more about triangle congruency theorems here:
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