Question

show that the bisectors of angles of a parllelogram from a rectangle​

Answer 1

Let ABCD be a parallelogram and P, Q, R and S be the points of intersection of the bisectors of ∠A and ∠D, ∠A and ∠B, ∠B and ∠C, ∠D and ∠C respectively of parallelogram ABCD.

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AP bisects ∠A and DP bisects ∠D.

Therefore ∠DAP + ∠ADP = 1/2

∠A + 1/2 ∠D = 1/2 (∠A + ∠D)

⇒ ∠DAP + ∠ADP

= 1212 x 180° = 90° (sum of consecutive ∠s of a parallelogram) ∠DAP + ∠ADP = 90° ….(i)

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Also in ∆ADP,

∠DAP + ∠ADP + ∠APD = 180° (angle sum property of a triangle) ⇒ 90° + ∠APD = 180°

⇒ ∠APD = 90° [using (ii)]

But ∠QPS = ∠APD (vertically opposite angles)

⇒ ∠QPS = 90° ..(ii)

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Similarly, BR bisects ∠B and CR bisects, ∠C,

therefore ∠QRS = 90° …(iii)

Also, AQ bisects ∠A and BQ bisects ∠B and DS bisects ∠D and CS bisects ∠C,

therefore ∠PQR = 90° …(iv)

and ∠PSR = 90° ..(v)

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So, using (ii), (iii), (iv) and (v),

we can say that PQRS is a quadrilateral in which all angles i.e. ∠P, ∠Q, ∠R and ∠S are right angles. Hence, PQRS is a rectangle.

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