ABCD is a parallelogram with ∠A = 80°. The internal bisectors of ∠B and ∠C meet each other at O. Find the measure of the three angles of ΔBCO.

Dear Student

∠A = 80°
We know that the opposite angles of a parallelogram are equal.
∠A = ∠C = 80°
And
∠OCB = (1/2) × ∠C
= (1/2) × 80°
= 40°
∠B = 180° – ∠A (the sum of interior angles on the same side of the transversal is 180)
= 180° – 80°
= 100°
Also,
∠CBO = (1/2) × ∠B
= (1/2) × 100°
= 50°
By the angle sum property of triangle BCO,
∠BOC + ∠OBC + ∠CBO = 180°
∠BOC = 180° – (∠OBC + CBO)
= 180° – (40° + 50°)
= 180° – 90°
= 90°
Hence, the measure of all the three angles of a triangle BCO is 40°, 50° and 90°.

Thanks