In the given figure, O is the center of the circle and XOY is a diameter. If XZ is any other chord of the circle, show that XY>XZ.
Given: XOY is a diameter and XZ is any chord of the circle.
To prove: XY>XZ
Proof:
In ∆XOZ,
OX + OZ > XZ … sum of any sides in a triangle is a greater than its third side
∴ OX + OY > XZ … As OZ = OY, radius of circle
Hence, XY > XZ …As OX + OY = XY
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