question_answer

\[APB\] and \[CQD\] are parallel lines and a transversal \[PQ\] cuts them at \[P\] and\[Q\]. Which of the following is formed by the bisectors of angles \[APQ,\,\,BPQ,\,\,CQP\] and\[PQD\]?

A)  A rectangle

B)         A rhombus

C)         A square

D)         Any other parallelogram

Correct Answer: A

Solution :

            \[\angle APQ+\angle CQP={{180}^{o}}\]             \[\Rightarrow \]\[\frac{\angle APQ}{2}+\frac{\angle CQP}{2}{{90}^{o}}\] \[\Rightarrow \]\[\angle RPQ+\angle PQR={{90}^{o}}\] \[\Rightarrow \]\[\Delta PQR\]is a right angled triangle Similarly, \[\Delta PQR\]is a right angled triangle \[\Rightarrow \]\[PQRS\]is rectangle.