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Question
Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.
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Solution
Given: `square`ABCD is a parallelogram.
To prove: `square`PQRS is a rectangle.
Proof:
`square`ABCD is a parallelogram. ...(Given)
∠ADC + ∠BCD = 180° ...(Adjacent angles of a parallelogram are supplementary.)
Multiplying each side by `1/2`,
`1/2` ∠ADC + `1/2` ∠BCD = `1/2xx180°` ...(i)
But,
`1/2` ∠ADC = ∠PDC ...(Ray DP bisects ∠ADC) ...(ii)
And `1/2` ∠BCD = ∠PCD ...(Ray CP bisects ∠BCD) ...(iii)
∴ ∠PDC + ∠PCD = 90° ...[From (i), (ii) and (iii)] ...(iv)
In ΔPDC,
∠PDC + ∠PCD + ∠DPC = 180° ...(The sum of the measures of the three angles of a triangle is 180°.)
∴ 90° + ∠DPC = 180° ...[from (iv)]
∴ ∠DPC = 180° – 90°
∴ ∠DPC = 90°
That means ∠SPQ = 90° ...(D-S-P, P-Q-C) ...(v)
Similarly, we can prove that, ∠SRQ = 90° ...(vi)
Similarly, ∠ASD = 90° and ∠BQC = 90° ...(vii)
∠PSR = ∠ASD ...(vertex angle)
∴ ∠PSR = 90° ...[From (vii)] ...(viii)
Similarly, ∠PQR = 90° ...(ix)
In `square`PQRS,
∠SPQ = ∠SRQ = ∠PSR = ∠PQR = 90° ...[From (v), (vi), (viii) and (ix)]
∴ `square`PQRS is a rectangle.
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