Question

In the adjoining figure AB > AC. If OB and OC are the bisectors of ∠ABC and ∠ACB, respectively, then prove that OB > ОС.

Answer 1

Given:

• In triangle ABC, AB > AC.
• OB and OC are the internal bisectors of ∠ABC and ∠ACB, respectively, so O is the intersection of those bisectors.

To Prove:

• OB > OC.

Proof:

1. From AB > AC,

The angle opposite the longer side is larger.

Hence ∠C > ∠B.

2. Since OB and OC are angle bisectors,

`∠OBD = 1/2 ∠ABC = B/2` where D is the foot of the perpendicular from O on BC,

`∠OCD = 1/2 ∠ACB = C/2`.

3. Let D be the perpendicular foot from O to BC; then OD = r the inradius and triangles OBD and OCD are right triangles at D.

4. In right triangle OBD,

`sin (B/2) = "Opposite"/"Hypotenuse"`

= `(OD)/(OB)`

= `r/(OB)` 

Hence, `OB = r/sin (B/2)`.

Similarly, in right triangle OCD,

`OC = r/sin (C/2)`

5. Because ∠C > ∠B, we have `C/2 > B/2`.

For angles in (0, 90°), the sine function is increasing.

So, `sin (C/2) > sin (B/2)`.

6. Thus, `1/sin (B/2) > 1/sin (C/2)` and multiplying by r > 0 gives `OB = r/sin (B/2) > r/sin (C/2) = OC`.

OB > OC, as required.