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Question

In the following diagram, the bisectors of interior angles of the parallelogram PQRS en-close a quadrilateral ABCD.

Show that: (i) PSB+SPB=90(ii) PBS=90 (iii) ABC=90(iv) ADC=90 (v) A=90(vi) ABCD is a rectangleThus, the bisectors of the angles of a parallelogram enclose a rectangle.

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Solution

Given : In parallelogram ABCD bisector ofangles P and Q, meet at A, bisectors of angleR.ABCD as shown in the figure.To Prove :(i) PSB+SPB=90(ii) PSB=90 (iii) ABC=90(iv) ADC=90 (v) A=9(vi) ABCD is a rectangle Proof :In parallelogram PQRS,PS || QR (opposite sides)P+Q=180 and AP and AQ are the bisectors of consecutive angle angleP and Q of the parallelogram. APQ+AQP=12×180=90But inΔAPQ,A+APQ+AQP=180 (Angles of a triangle) A+90=180 A=18090(v) A=90Similarly PQ || SR(i) PSB+SPB=90(ii) and PBS=90But, ABC=PBS(Vertically opposite angles)(iii) ABC=90Similarly we can prove that(iv) ADC=90and C=90(v) ABCD is a rectangle(Each angle of a quadrilateral is 90) Hence proved.


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