Question

prove that 1/BD^2 = 1/AB^2 + 1/BC^2​

Answer 1

Answer:

Step-by-step explanation:

Triangle ABC is right angled at C.Let BC = a, CA = b, AB = c.

(i) Area of ΔABC = 1/2 × Base × Height = 1/2 × BC × AC = 1/2 ab

   Area of ΔABC = 1/2 × Base × Height = 1/2 × AB × CD = 1/2 cp

   ⇒ 1/2 ab = 1/2 cp

   ⇒ ab = cp

   Hence proved.

(ii) In right angled triangle ABC,

AB2 = BC2 + AC2

c2 = a2 + b2

(ab/p)2 = a2 + b2

a2b2/p2 = a2 + b2 -------- From proof (1)

1/p2 = (a2 + b2) / a2b2

1/p2 = (a2 / a2b2 + b2/ a2b2)

1/p2 = (1/b2 + 1/a2)

1/p2 = (1/a2 + 1/b2)

Hence proved.