Question

∆ ABC is a triangle were angle C=90 degree. Let BC=a, CA=b, AB=c and let 'p' be the length of perpendicular from C on AB. Prove that
(1) cp=ab,
(2) 1 1 1
--- = --- = ---
p^2 a^2 b^2 .

Answer 1

1)Area of ΔABC = 1/2 × B × H = 1/2 × BC × AC = 1/2 ab
    Area of ΔABC = 1/2 × B × H = 1/2 × AB × CD = 1/2 cp
    ⇒ 1/2 cp = 1/2 ab
As the half of cp and ab are equal,so we can say that;
       cp = ab
    Hence,cp = ab is proved.

2)In  ∆ABC,
AB^2 = BC^2 + AC2
c^2 = a^2 + b^2
(ab/p)^2 = a^2 + b^2
a^2b^2/p^2 = a^2 + b^2
1/p^2 = (a^2 + b^2) / a^2b^2
1/p^2 = (a^2 / a^2b^2 + b^2/ a^2b^2)
1/p^2 = (1/b^2 + 1/a^2)
1/p^2 = (1/a^2 + 1/b^2)

Hope it helps!!!

Answer 2

I sent you images . Hope, it helps you. (•‿•)