question_answer

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form

A)  Kite                             

B)         Rhombus                     

C)         Rectangle                     

D)         Trapezium

Correct Answer: C

Solution :

Given, APB and CQD are two parallel lines. Let the bisectors of angles APQ and CQP meet at a point M and bisectors of angles BPQ and PQD meet at a point N. Since, \[APB||CQD\] \[\therefore \]\[\angle APQ=\angle PQD\] [Alternate interior angles] \[\Rightarrow \]\[\frac{1}{2}\angle APQ=\frac{1}{2}\angle PQD\] \[\Rightarrow \]\[\angle MPQ=\angle NQP\] This shows that alternate interior angles are equal. \[\therefore \]      \[PM||QN\] Similarly,\[\angle NPQ=\angle MQP,\]which shows that alternate interior angles are equal. \[\therefore \]      \[PN||QM\] So, quadrilateral PMQN is a parallelogram. Also, \[\angle CQP+\angle DQP={{180}^{o}}\]  [Linear pair] \[\Rightarrow \]\[2\angle MQP+2\angle NQP={{180}^{o}}\] \[\Rightarrow \]\[2(\angle MQP+\angle NQP)={{180}^{o}}\] \[\Rightarrow \]\[\angle MQN={{90}^{o}}\] Thus, PMQN is a rectangle.