question_answer

Sides BA and DC of a quadrilateral ABCD are produced as shown in figure, then \[\angle \mathbf{a}+\angle \mathbf{b}\] is equal to

A)  \[\angle x-\angle y\]                   

B)  \[\angle x+\angle y\]

C)  \[\angle x+\angle 2y\]                

D)  none of these

Correct Answer: B

Solution :

(b): Join BD: \[\angle \text{1}+\angle 2=\angle y\] \[\angle 3+\angle 4=\angle x\]             In \[{{\Delta }^{1e}}\] ABD \[\angle b=\angle 1+\angle 3\] (exterior angle= sum of interior \[\angle s\]) Similarly, in \[{{\Delta }^{1e}}BCD,\] \[\angle a=\angle 2+\angle 4\] Adding \[\angle b+\angle a=\angle 1+\angle 2+\angle 3+\angle 4\] \[\Rightarrow \angle a+\angle b=\angle x+\angle y.\]