Question

Diagonals of a quadrilateral ABCD bisect each other. If $\angle A={45}^{\circ }$, then $\angle B=$

• A. ${115}^{\circ }$
• B. ${120}^{\circ }$
• C. ${125}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is D

${135}^{\circ }$

Diagonals AC and BD of quadrilateral ABCD bisect each other at O.

$\because AO=OC,BO=OD\therefore ABCDisa||gm.\angle A={45}^{\circ }But\angle A+\angle B={180}^{\circ }\left(\text{Sum of consecutive angles}\right)\therefore \angle B={180}^{\circ }-\angle A={180}^{\circ }-{45}^{\circ }={135}^{\circ }$