ABC is right triagle right angled at C Let BC=a, CA=b,AB=c and let p be the lenght of perpendicular from C on AB. Prove that 1/p square = 1/a square +1/b square

 

Area of triangle ACB with base AB = (1/2)AB*CD = (1/2)cp

Area of triangle ACB with base AC = (1/2)AC*BC = (1/2)ba

 

Both areas are same. Therefore, on equating, we get,

c = ab/p           ... (1)

 

Now, using Pythagoras theorem in triangle ABC, we have

AB² = AC²+BC²

c² = b²+a²

a²b²/ p² = b²+a²         [Using (1)]

1/ p²= (b²+a²)/ a²b²

Hence, 1/ p²= 1/a²+1/b²