Prove that among all the chords of a circle passing through a given point inside the circle that one is smallest which is perpendicular to the diameter passing through the point.
Answers
Let P be the given point inside a circle with centre O.
Draw the chord AB which is perpendicular to the diameter XY through P.
Let CD be any other chord through P. Draw ON perpendicular to CD from O.
Then, △ONP is a right triangle.
Therefore, its hypotenuse OP is larger than ON.
We know that the chord nearer to the centre is larger than the chord which is farther from the centre.
Therefore, CD>AB
In other words, AB is the smallest of all chords passing through P.
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