Question

apollonius theorem proof


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Answer 1

Choose the origin of the rectangular form of the

Cartesian coordinates at the point O and the x-axis coming along the sides MN and also OY as y – axis. If in case MN = 2a, then the coordinates of the points M, as well as N, are (a, 0) and (- a, 0) respectively. If coordinates of the point L are (b, c), then

LO² = (C – 0)² + (b – 0)² , (Since the coordinates of the point O are {0, 0})

= c² + b²;

LM² = (c – 0)² + (b + a) ² = c² + (a + b)²

MO² = (0 – 0)² + (- a – 0)² = a²

also, LN² = (c – 0) ² + (b – a) ² = c² + (a – b)²

Therefore, LN² + LM² = c² + (a + b) ² + c² + (b – a)²

= 2c² + 2 (a² + b²)

= 2(b² + c²) + 2a²

= 2LO² + 2MO²

= 2 (LO² + MO²).

= 2(MO² + LO²). {Hence Proved}

Answer 2

Answer:

Let us choose origin of rectangular Cartesian co-ordinates at O and x-axis along the side MN and OY as the y – axis . If MN = 2a then the co-ordinates of M and N are (- a, 0) and (a, 0) respectively. Referred to the chosen axes if the co-ordinates of L be (b, c) then

LO² = (b - 0)² + (C - 0)² , [Since, co- ordinates of O are (0, 0)]

= b² + c²;

MO² = (- a - 0)² + (0 – 0)² = a²

LM² = (b + a) ² + (c – 0)² = (a + b)² + c²

And LN² = (b - a) ² + (c - 0) ² = (a - b)² + c²

Therefore, LM² + LN² = (a + b) ² + c² + (b - a)² + c²

= 2(a² + b²) + 2c²

= 2a² + 2(b² + c²)

= 2MO² + 2LO²

= 2(MO² + LO²).

= 2(LO² + MO²). Proved.