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In triangle ABC, if BD is the bisector of angle B and segment DE is drawn parallel to side AB with ∠BDE = 28°, what are the measures of angles B and C?

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Final answer:

Angle B is 56°, being twice the measure of ∠BDE due to BD being the bisector. Angle C is 96°, calculated from the sum of angles in a triangle being 180°.

Step-by-step explanation:

To find the measures of angles B and C in triangle ABC, given that BD is the bisector of angle B and segment DE is parallel to side AB with ∠BDE = 28°, we need to use the properties of parallel lines and angle bisectors in a triangle.

Since DE is parallel to AB, and BD is a transversal, the alternate interior angles are equal. Therefore, ∠ABD = ∠BDE, which is 28°. Because BD is the bisector of angle B, this implies that the full measure of angle B is 2×∠ABD, which is 56°.

To find angle C, we use the fact that the sum of angles in a triangle is 180°. Therefore, ∠C = 180° - ∠B - ∠A. Since ∠B is 56° and ∠A = ∠BDE (because of parallel lines) which is 28°, we get ∠C = 180° - 56° - 28° which is 96°.

User Mike Steinert
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