In ∆ABC, O and I are circumcentre and indenter of ∆ respectively and R and r are the respective radii of circum circle and in circle.
To prove (IO)2 = R2 – 2Rr
First of all we will find IO. Using cosine law in ∆AOI
cos ∠OAI = OA2 + AI2 – OI2/2. OA. AI . . . . . . . . . . . . . . . . (1)
where OA = R
In ∆ AID, sin A/2 = r/AI
AI = r/sin A/2
= 4R sin B/2 sin C/2 [Using r = 4R sin A/2 sin B/2 sin C/2]
Also, ∠OAI = ∠IAE - ∠OAE
= A/2 – (90° - ∠AOE)
= A/2 - 90° + 1/2 ∠AOC
= A/2 - 90° + 1/2. 2B (∵ O is circumcentre ∴ ∠AOC = 2∠B)
= A/2 + B – A + B + C/2
= B – C/2
Substituting all these values in equation (1) we get
Cos (B – C)/2 = R2 + 16 R2 sin2 B/2 sin2 c/2 – OI2 | 2.R.4 R sin C/2
⇒ OI2 = R2 + 16R2 sin2 B/2 sin2 C/2 – 8R2 sin B/2 sin C/2 cos B – C/2
= R2 [1 + sin B/2 sin C/2{2 sin B/2 sin C/2 – cos B – C/2 }]
= R2 [1 + 8 sin B/2 sin C/2 {2 sin B/2 sin C/2 – cos B/2 cos C/2 – sin B/2 sin C/2}]
= R2 [1 + 8 sin C/2 { sin B/2 sin C/2 – cos B/2 cos C/2]
= R2 [1-8 sin B/2 sin C/2 cos B + C/2]
= R2 [1 – 8 sin A/2 sin B/2 sin C/2]
= R2 – 2R. 4R sin A/2 sin B/2 sin C/2
= R2 – 2Rr. Hence Proved
Again if ∆ OIB is right ∠ed ∆ then
⇒ OB2 = OI2 + IB2
⇒R2 = R2 – 2Rr + r2/sin2 B/2
NOTE THIS STEP:
[∵ In ∆ IBD sin B/2 = r/IB]
⇒ 2R sin2 B/2 = r
⇒ 2abc/4∆ (s – a) (s – c)/ac = ∆/s
⇒ b(s – a) (s – c) = 2 (s – a)(s – b)(s – c)
⇒b = 2s – 2b ⇒ b = a + b – c
⇒ a + c = 2b ⇔ a, b, c are in A. P.
⇒ b is A. M> between a and c. Hence Proved.s
