Question

If ABCD is parallelogram. E is mid-point of AB and CE bisects $\angle BCD$, then $\angle DEC$ is _____.

• A. ${60}^o$
• B. ${90}^o$
• C. ${100}^o$
• D. ${120}^o$

Answer 1

The correct option is B ${90}^o$

As ABCD is parallelogram and E is mid-point of AB.
& CE bisects $\angle BCD$.
So, DE will bisect $\angle ADC$
As ABCD is parallelogram,
So, $\angle ADC$= $\angle BCD$=${90}^o$
So, $\angle EDC$= $\angle ECD$=${90}^o/2$=${45}^o$
In triangle CDE,
$\angle EDC+\angle ECD+\angle DEC={180}^o$
${45}^o+{45}^o+\angle DEC={180}^o$
$\angle DEC={180}^o-{90}^o={90}^o$
Answer (B) ${90}^o$