In the given figure, ΔABC is right- angled at C. Let BC = a, CA = b, AB = c and CD = p, where CD 1p2 = 1a2 + 1b2
(i) Area of ΔABC = 12AB × CD = 12cp.
Also, area of ΔABC = 12BC × AC = 12ab.
∴12cp = 12ab ⇒ cp = ab
(ii) cp = ab ⇒ p = abc
⇒ p2 = a2b2c2
⇒ 1p2 - c2b2b2 - a2+b2a2b2 [ ∵ c2 = a2 + b2]
⇒1p2 - 1a2 + 1b2