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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∘=90∘But inΔAPQ,∠A+∠APQ+∠AQP=180∘ (Angles of a triangle)⇒ ∠A+90∘=180∘ ⇒ ∠A=180∘−90∘(v) ∠A=90∘Similarly PQ || SR(i) ∴ ∠PSB+SPB=90∘(ii) and ∠PBS=90∘But, ∠ABC=∠PBS(Vertically opposite angles)(iii) ∴ ∠ABC=90∘Similarly we can prove that(iv) ∠ADC=90∘and ∠C=90∘(v) ∴ ABCD is a rectangle(Each angle of a quadrilateral is 90∘) Hence proved.

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