Question

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.​

Answer 1

Given,

∠A = 80°

We know that ,

the opposite angles of a parallelogram are equal.

$\sf{∠A = ∠C = 80°}$

And

$\sf{∠OCB =(\frac{ 1}{2}) × ∠C}$

$\sf{=(\frac{ 1}{2}) × 80°}$

$\sf{= 40°}$

∠B = 180° – ∠A (the sum of interior angles on the same side of the transversal is 180)

$\sf{= 180° – 80°}$

$\sf{= 100°}$

Also,

$\sf{∠CBO = (\frac{1}{2}) × ∠B}$

$\sf{=(\frac{1}{2}) × 100°}$

= 50°

By the angle sum property of triangle BCO,

$\sf{∠BOC + ∠OBC + ∠CBO = 180°}$

$\sf{∠BOC = 180° – (∠OBC + CBO)}$

$\sf{= 180° – (40° + 50°)}$

$\sf{= 180° – 90°}$

$\sf{= 90°}$

Hence, the measure of all the three angles of a triangle BCO is 40°, 50° and 90°