MHB Proving angles are equal in a triangle in a circle.

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To prove that angles CAB and CBA are equal in a triangle inscribed in a circle, the discussion highlights the use of the inscribed angle theorem and the alternate segment theorem. By connecting the angles to the measures of arcs AC and CB, it is established that these arcs are equal, leading to the conclusion that angles CAB and CBA are congruent. The alternate segment theorem is referenced to show that the angle between a tangent and a chord equals the angle in the alternate segment, reinforcing the isosceles nature of triangle CBA. Thus, the proof demonstrates the equality of the angles based on the properties of circles and triangles. The discussion effectively illustrates the geometric relationships involved in the proof.
markosheehan
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hi can someone help me work this out. i think it has something to do with the exterior angle of a triangle is equal to the sum of the interior angles but i can't work it.
 

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Prove that $\angle CAB=\angle CBA$ by connecting these angles to measures of arcs using the inscribed angle theorem and its corollary about tangent lines.
 
how do you know the arcs AC and CB are the same
 
Last edited:
markosheehan said:
how do you know the arcs AC and CB are the same

Are you familiar with the Alternate Segment theorem, which states that

The angle between the tangent and chord at the point of contact is equal to the angle in the alternate segment.

View attachment 6099

So now do you see a similarity between the angle marked in blue in the above diagram & the angle marked in blue in your diagram.

$\therefore$ It can be said that $\angle CAT = \angle CBA$ using alternate segment theorem

Now what can be said about the triangle CBA using it's angles?
 

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It's isosceles
 

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