Question

Abc is a right angle triangle in which angle c is = 90 . If p is the length of the perpendicular From C to AB = c , BC = a and CA = b , then prove that 1/p square = 1/a square+ 1/b square

Answer 1

say AB=x
then
a^2+b^2=x^2

and x=(a^2-c^2)^1/2+(b^2-c^2)^1/2
sqaring,
x^2=a^2+b^2-2*c^2+2*(a^2-c^2)^1/2*(b^2-c^2)^1/2

putting, a^2+b^2=x^2
we get
c^2=(a^2-c^2)^1/2*(b^2-c^2)^1/2
sqaring
c^4=(ab)^2-(ac)^2-(bc)^2+c^4

(ab)^2=(ac)^2+(bc)^2
dividing by (abc)^2,

1/c^2=1/a^2+1/b^2

hence proved