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Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. |
Answer» Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of `angleA, angleB, angleC and angleD,` respectively. To prove Quadrilateral PQRS is a rectangle. Proof Since, ABCD is a parallelogram, then DC||AB and DA is a transversal. We have, `" "angleA+angle D= 180^(@)` `" "` [sum of cointerior angles of parallelogram is `180^(@)`] `rArr" "(1)/(2)angleA+(1)/(2)angleD=90^(@)" "`[dividing both sides by 2] `rArr" "anglePAD+anglePDA=90^(@)` ltBrgt `rArr" "angleAPD=90^(@)" "` [since, sum of all angles of a triangle is `180^(@)`] `therefore" "angleSPQ=90^(@)" "` [vertically opposite angles] Similarly, `" "anglePQR=90^(@)` `" "angleQRS=90^(@)` and `" "anglePSR=90^(@)` Thus, PQRS is a quadrilateral whose each angle is `90^(@)`. Hence, PQRS is a rectangle. `" "` Hence proved. |
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