A) \[\angle CED\]
B) \[\frac{1}{2}\angle CED\]
C) \[2\angle CED\]
D) None of these
Correct Answer: C
Solution :
In quadrilateral ABCD, (angles sum property.) \[\angle A+\angle B+\angle C+\angle D={{360}^{o}}\] \[\Rightarrow \]\[\angle A+\angle B+2\angle 2+2\angle 1={{360}^{o}}\] \[(\because \,\angle 2=\frac{1}{2}\angle C\,\text{and}\,\angle 1=\frac{1}{2}\angle D)\] \[\Rightarrow \]\[\angle +\angle B={{360}^{o}}-2(\angle 1+\angle 2)\] ?(i) In \[\Delta DEC,\] \[\angle 1+\angle 2+\angle CED={{180}^{o}}\] (angle sum property) \[\Rightarrow \]\[\angle 1+\angle 2={{180}^{o}}-\angle CED\] ?(ii) From (i) and (ii), we get \[\angle A+\angle B={{360}^{o}}-2({{180}^{o}}-\angle CED)\] \[\angle A+\angle B={{360}^{o}}-{{360}^{o}}+2\angle CED\] \[\Rightarrow \]\[\angle A+\angle B=2\angle CED\]
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