Q. 133.8( 6 Votes )

# The circumcenter of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90°.

Answer :

Let ABC be the triangle whose circumcenter is O.

∠OBC = ∠OCB = θ (**opposite angles of equal sides**

**In** **Δ****BOC, using the angle sum property of tringle, sum of all angles is 180°, we have:****∠****BOC +** **∠****OBC +** **∠****OCB = 180°** **⇒** **∠****BOC +** **θ + θ** **= 180°**

**⇒** **∠****BOC = 180° -2θ**

**Also, in a circle, angle subtended by an arc at the center is twice the angle subtended by it at any other point in the remaining part of the circle.** **∠****BOC = 2****∠****BAC**

**⇒** **∠****BAC = 1/2(****∠****BOC)**

**⇒** **∠****BAC = 1/2(****180° - 2θ****)**

**⇒** **∠****BAC = (****90° - θ****)**

**⇒** **∠****BAC + θ = 90°**

**⇒** **∠****BAC +** **∠****OBC = 90°**

Hence, proved.

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