Question

In the adjoining figure OB and OC are the bisectors of ∠DBC and ∠ECB, respectively. If AC > AB prove that OB > OC.

Answer 1

Given: A, B, C, D, E, O are as in the figure. OB bisects ∠DBC and OC bisects ∠ECB. AC > AB.

To Prove: OB > OC.

Proof [Step-wise]:

1. In triangle ABC,

AC > AB

⇒ ∠ABC > ∠ACB

2. Since D lies on the extension of BA beyond B, ∠DBC and ∠ABC are supplementary. 

So, ∠DBC = 180° – ∠ABC. 

Similarly ∠ECB = 180° − ∠ACB.

3. OB bisects ∠DBC,

So, `∠OBC = (1/2) ∠DBC = 90^circ - (1/2) ∠ABC`.

OC bisects ∠ECB,

So, `∠OCB = (1/2) ∠ECB = 90^circ - (1/2)∠ACB`.

4. From step 1, 

∠ABC > ∠ACB 

⇒ `(1/2) ∠ABC > (1/2) ∠ACB` 

Hence, `90^circ - (1/2) ∠ABC < 90^circ - (1/2) ∠ACB`.

Therefore, ∠OBC < ∠OCB.

5. In triangle BOC, the larger side is opposite the larger angle.

Here, ∠OCB > ∠OBC.

So, the side opposite ∠OCB (which is OB) is greater than the side opposite ∠OBC (which is OC).

Therefore, OB > OC.