Q. 43

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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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