MHB Proving Inscribed Angle Theorem for Triangle Side Lengths

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The discussion centers on proving the Inscribed Angle Theorem using triangle OAB, which is identified as isosceles. The angles at the base of the triangle are derived from the relationship between the central angle BOC and the inscribed angle BAC, where BAC is half of BOC. It is established that since BOC measures 30 degrees, BAC must be 15 degrees. Participants seek clarification on the isosceles nature of the triangle and the reasoning behind the angle measurements. The discussion emphasizes the geometric properties of inscribed angles in relation to central angles.
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I already found the side lengths that's being asked in the problem. But the part that I'm having difficulty is the Proving part. Can you help me do that? Thanks!
 

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Triangle $OAB$ is isosceles, so what must the other two angles be?
 
There is another way of seeing this that proves in one single argument.

Notice that $BOC$ is the angle seen from the center of the arc, while $BAC$ is an angle seen from the circumference covering the same arc. This means that $BAC$ is half the angle $BOC$, therefore $BAC = 15^{\circ}$.

Sorry for barging in. :)

Best wishes.
 
MarkFL said:
Triangle $OAB$ is isosceles, so what must the other two angles be?

how did you know that it's an isosceles triangle?

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Fantini said:
There is another way of seeing this that proves in one single argument.

Notice that $BOC$ is the angle seen from the center of the arc, while $BAC$ is an angle seen from the circumference covering the same arc. This means that $BAC$ is half the angle $BOC$, therefore $BAC = 15^{\circ}$.

Sorry for barging in. :)

Best wishes.

I didn't get it. Can you please elaborate?
 
paulmdrdo said:
how did you know that it's an isosceles triangle?

$\overline{OA}$ and $\overline{OB}$ are both radii of the same circular segment, right?
 
The problem says that $ABC$ is the arc of a circumference. This means that $BAC$ is an inscribed angle in this circumference. See here.

Now look at the angle $BOC$. This is the central angle. According to the inscribed angle theorem this means that the angle $BAC$ is half the angle $BOC$. Since $BOC = 30^{\circ}$ then $BAC = 15^{\circ}$.
 
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