Q12 of 24 Page 157

C is the midpoint of the line segment AB and PQ is any straight line. The minimum distances of the line PQ from the points A, B and C are AR, BS and CT respectively; let us prove that, AR + BS = 2CT.


As AR, BS, CT are the minimum distances of the line PQ from the points A, B and C, this can be only achieved when AB||PQ and AR, CT and BS are perpendicular to it.


This makes ARTC and CTSB as a parallelogram.


⇒ CT = AR from ARTC parallelogram,


And CT = BS from CTSB parallelogram.


On adding the above two equations, we get


⇒ AR + BS = 2CT


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