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Asked by Suri Last Modified
Answer 
Pooja R. Jain
Let's consider the parallelogram ABCD with ∠A = 80°. Since ABCD is a parallelogram, opposite angles are equal. Therefore, ∠B = 80°.
Now, let I1 be the internal bisector of ∠B and I2 be the internal bisector of ∠C. The point where these bisectors meet, denoted as O, is the incenter of triangle BCD.
Now, ∠BCO is half of ∠B, and ∠CBO is half of ∠C.
∠BCO=12⋅∠B=12⋅80°=40°∠BCO=21⋅∠B=21⋅80°=40°
∠CBO=12⋅∠C∠CBO=21⋅∠C
Since ∠C and ∠A are opposite angles in the parallelogram, ∠C = ∠A = 80°. Therefore,
∠CBO=12⋅80°=40°∠CBO=21⋅80°=40°
Finally, since the sum of angles in a triangle is 180°, we can find ∠OBC:
∠OBC=180°−∠BCO−∠CBO=180°−40°−40°=100°∠OBC=180°−∠BCO−∠CBO=180°−40°−40°=100°
So, the measures of the angles of triangle BCO are 40°40°, 40°40°, and 100°100°.
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