ABCD is a parallelogram, E is the mid – point of AB and CE bisects ∠ BCD. Then ∠ DEC is

Given: ABCD is a parallelogram and E is the mid – point of AB such that CE bisects ∠ BCD.

construction: Join DE. The figure is given below:

Let ∠ BCD = 2x

⇒ ∠ BCE = ∠ ECD = ∠ BCD/2

Now, let ∠ ADC = 2y

And as we joined DE it will also bisect ∠ ADC.

Therefore, ∠ EDC =

Now, as we know sum of adjacent angles of parallelogram is equal to 180°.

⇒ ∠ BCD + ∠ ADC = 180°

⇒ 2x + 2y = 180°

⇒ 2 (x + y) = 180°

⇒ x + y = 90°

Now, In triangle CED,

∠ CED + ∠ EDC + ∠ DCE = 180°

⇒ ∠ CED + x + y = 180°

⇒ ∠ CED + 90° = 180°

⇒ ∠ CED = 180° – 90°

⇒ ∠ CED = 90°

Hence required angle is 90°.