Question

IN triangle abc prove That ab-r1r2/r3=bc-r2r3/r1=ca-r3r1/r2=r​

Answer 1

In triangle ABC, prove that (ab - (r1r2))/r3 = (bc - (r2r3))/r1 = (ca - (r3r1))/r2 = r

Given,

A triangle ABC

To prove,

(ab - (r1r2))/r3 = (bc - (r2r3))/r1 = (ca - (r3r1))/r2 = r

Solution,

Since a = 2 R sin A and b = 2 R sin B,

(ab - (r1r2))/r3 = ((2 R sin A * 2 R sin B) - r1r2) / r3

and, r1 = 4 R sin A/2 cos B/2 cos C/2

r2 = 4 R cos A/2 sin B/2 cos C/2

r3 = 4 R cos A/2 cos B/2 sin C/2

Then, (ab - (r1r2))/r3

= [(4R)^2 (sin A/2 cos B/2 cos C/2) - [(cos A/2 sin B/2 cos C/2) * (cos A/2 cos B/2 sin C/2)]] / 4 R cos A/2 cos B/2 sin C/2

= [4 R (1 - cos^2 (c/2)) sin A/2 sin B/2] / [sin (c/2)]

= [4 R (sin^2 (c/2)) sin A/2 sin B/2] / [sin (c/2)]

= 4 R sin (A/2) sin (B/2) sin (C/2)

= r

Similarly, we can prove that (bc - (r2r3))/r1 = r

and (ca - (r3r1))/r2 = r

Therefore, (ab - (r1r2))/r3 = (bc - (r2r3))/r1 = (ca - (r3r1))/r2 = r

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