Triangle ABC is right angled at C. If p is the length of the perpendicular from C to AB and a, b, c are the lengths of the opposite angles A, B and C respectively then prove that 1/p^2 = 1/a^2 + 1/b^2.

Solution: area of triangle ABC = ½ * base * height

= ½ * AB * CD

= ½ (c * p)                               …… (1) 

Also the area of triangle ABC can also be equal to,

= ½ * base * height

= ½ * BC * AC

= ½ (a * b)                               …… (2) 

From eq(1) and (2), we get

½ cp = ½ ab

= p = ab/c

= 1/p = c/ab

Squaring both sides, we get

\[\frac{1}{p^2}=\frac{c^2}{a^2b^2}=\frac{b^2+a^2}{a^2b^2}\]

\[\frac{1}{p^2}=\frac{b^2}{a^2b^2}=\frac{a^2}{a^2b^2}\]

\[\frac{1}{p^2}=\frac{1}{a^2}=\frac{1}{b^2}\]