question_answer

If the sides BA and DC of quadrilateral  ABCD are produced as shown in the given figure, then

A)                     \[x+y=a+b\]                

B)        \[x-y=a-b\]       

C)        \[\frac{x-y}{2}=a-b\]                

D)        \[2(x+y)=a+b\]

Correct Answer: A

Solution :

Join BD. In \[\Delta \Alpha \Beta D,\]we have \[\angle ABD+\angle ADB=b\]                                    ?(i) In\[\Delta CBD,\] we have \[\angle CBD+\angle CDB=a\]    ?(ii) Adding (i) and (ii), we get \[(\angle ABD+\angle CBD)+(\angle ADB+\angle CDB)\]\[=a+b\] \[\Rightarrow \]\[x+y=a+b\]