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  • If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form (A) a square (B) a rhombus (C) a rectangle (D) any other parallelogram

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

(A) a square                                      (B) a rhombus
(C) a rectangle                               (D) any other parallelogram

Answers (1)

best_answer

Answer:  (C) a rectangle
Solution.
Given : APB and COD are two parallel lines.
Construction : Let us draw the bisectors of the angles APQ, BPQ, CQP and PQD

Let the bisectors meet at point M and N
Since APB\parallel CQD
\angle APQ=\angle PQD                                   (Alternate angles)
and   \angle MPQ=\angle PQN                        (Alternate interior angles)
\therefore PM\parallel QN
Similarly  \angle BPQ=\angle PQC                (Alternate angles)
\Rightarrow PN\parallel QM
So, quadrilateral PMQN is parallelogram
Since CQD is a line
\therefore \angle CQD=180^{\circ}
\angle CQP+\angle PQD=180^{\circ}
2\angle MQP+2\angle NQP=180^{\circ}
2\left (\angle MQP+\angle NQP \right )=180^{\circ}
\angle MQP+\angle NQP=\frac{180}{2}
\Rightarrow \angle MQN=\frac{180}{2}
\Rightarrow \angle MQN=90^{\circ}
Hence, PMQN is a rectangle.
Therefore option (C) is correct.

 

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infoexpert26

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