Question

a circle with center o circumscribes triangle ABC. Prove that the angles in the three segments exterior to triangle ABC are together equal to four right angles.​

Answer 1

Answer:

From the original △ ABC:

m ∠ BAC + m ∠ ABC + m ∠ BCA = 180.

Since O is the center of the circumscribed circle OA = OB = OC (all radii of a circle are congruent). Therefore, there are three isosceles triangles ( △ OAB, △ OBC, and △ OAC). In an isosceles triangle, the base angles are congruent (as marked and labeled in the diagram).

m ∠ BAC = β−α

m ∠ ABC = β+θ

m ∠ BCA = θ−α

Substituting into the triangle angle sum equation:

( β−α)+(β+θ)+(θ−α)=180

Combining like terms:

2β+2θ−2α=180

Dividing by 2:

β+θ−α=90

Reordering:

θ+(β−α)=90

Substituting:

m ∠ OBC + m ∠ BAC = 90

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