Question

The sides BA and DC of quadrilateral ABCD are produced as shown. Then which of following correct

• A. $\text{2}{\text{x}}^{\text{o}}\text{+}{\text{y}}^{\text{o}}\text{=}{\text{a}}^{\text{o}}\text{+}{\text{b}}^{\text{o}}$
• B. ${\text{x}}^{\text{o}}\text{+}\frac{\text{1}}{\text{2}}{\text{y}}^{\text{o}}\text{=}\frac{{\text{a}}^{\text{o}}\text{+}{\text{b}}^{\text{o}}}{\text{2}}$
• C. ${\text{x}}^{\text{o}}\text{+}{\text{y}}^{\text{o}}\text{=}{\text{a}}^{\text{o}}\text{+}{\text{b}}^{\text{o}}$
• D. ${\text{x}}^{\text{o}}\text{+}{\text{a}}^{\text{o}}\text{=}{\text{y}}^{\text{o}}\text{+}{\text{b}}^{\text{o}}$

Answer 1

The correct option is C ${\text{x}}^{\text{o}}\text{+}{\text{y}}^{\text{o}}\text{=}{\text{a}}^{\text{o}}\text{+}{\text{b}}^{\text{o}}$

REF.Image

Solution :-

we will join BD,

$\therefore$ In $△ABD$

$\angle b=\angle a+\angle b$ __ (1) [external angle is equal to sum of the other two angles]

In $△CBD$

$\angle CBD+\angle CDB=a$ __ (2)

(1) + (2)

$\angle ABD+\angle AOB+\angle CBD+\angle CDB)=a+b$

$\Rightarrow x+y=a+b$