Prove that the line segment joining the point of contact of two parallel tangents to a circle is a diameter of a circle
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Let, XBY and PCQ be two parallel tangents to a circle with centre O.
Construction :
Join OB and OC
Draw OA || XY
Proof :
Now,
XB || AO
⟹ ∠ XBO + ∠ AOB = 180° ─━⧼ sum of adjacent interior angles is 180°
Now,
∠ XBO = 90° ─━⧼ a tangent to a circle is perpendicular to the radius through a point of contact
⟹ 90° + ∠ AOB = 180°
∴ ∠ AOB = 90°
-lly, ∠ AOC = 90°
Hence,
∠ AOB + ∠ AOC = 90° + 90° = 180°
Hence, BOC is a straight line passing through O.
Thus, line segment joining the points of contact of two parallel tangents of a circle passes through its centre.
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