Prove that a chord which is not passing from centre is less than diameter?

Answer 1

Suppose AB is a chord of a circle which does not pass through its centre.From A, draw the diameter AC. Join BC.

Now, by the Circle Theorems (different courses number them differently), angle ABC is a right angle (angle subtended by a semicircular arc).

In other words, ABC is a right angled triangle with its right angle at B.

If |AB| represents the length of the line AB, and so on,

by Pythagoras's theorem, |AC|^2 = |AB|^2 + |BC|^2

then since |BC|^2 > 0, |AC|^2 > |AB|^2