Question

In figure, side AB and AC are produced to point D and E respectively. Bisectors BO and CO of ∠DBC and ∠ECB respectively meet at O. If AB > AC. Prove that OC > OB.

Answer 1

Given: ABC is a triangle in which AB and AC is produced up to D and E respectively and bisectors of ∠DBC and ∠ECB meet at O. Also AB > AC.

To prove: OC > OB

Proof: In ΔABC

AB > AC

⇒ ∠ACB > ∠ABC …(i)

Also BO and CO are the bisectors of ∠DBC and ∠BCE respectively.

∠OBD = ∠OBC

and ∠OCB = ∠OCE

∠ACB = 180° – ∠BCE

⇒ ∠ACB = 180° – 2∠OCB …(ii)

Similarly ∠ABC = 180° – 2 ∠OBC…(iii)

From (i), (ii) and (iii), we have

180° – 2∠OCB > 180° – 2∠OBC

⇒ – 2∠OCB > – 2∠OBC

⇒ ∠OBC > ∠OCB

⇒ OC > OB

Hence proved.