Sun August 26, 2012 By: Dhilip N

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

Expert Reply
Sun August 26, 2012
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
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