Mathematics
AB and CD are two equal chords of a circle with center O which intersect each other at right angle at point P. If OM ⊥ AB and ON ⊥ CD; show that OMPN is a square.
Circles
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Answer
Since, OM ⊥ AB and ON ⊥ CD.
∴ ∠OMP = ∠ONP = 90°.
Given,
AB and CD intersect at right angle.
∴ ∠MPN = 90°.
We know that,
Sum of angles of quadrilateral equal to 360°.
∴ ∠OMP + ∠ONP + ∠MPN + ∠MON = 360°
⇒ 90° + 90° + 90° + ∠MON = 360°
⇒ 270° + ∠MON = 360°
⇒ ∠MON = 360° - 270° = 90°.
Given,
AB = CD
We know that,
Equal chords are equidistant from the center.
∴ OM = ON = x (let) …………(1)
Since,
⇒ AB ⊥ CD and ON ⊥ CD
∴ AB || ON
∴ MP || ON.
⇒ OM ⊥ AB and CD ⊥ AB
∴ OM || CD
∴ OM || PN.
Since, in quadrilateral OMPN opposite sides are parallel.
∴ OMPN is parallelogram.
We know that,
Opposite sides of parallelogram are equal.
∴ MP = ON = x and OM = PN = x
∴ MP = ON = OM = PN.
Since, all sides of quadrilateral OMPN are equal and each interior angle to 90°.
Hence, proved that OMPN is a square.
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