MHB Proving Inscribed Angle Theorem for Triangle Side Lengths

[paulmdrdo1]

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!

 

[MarkFL]

Triangle $OAB$ is isosceles, so what must the other two angles be?

 

[Fantini]

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.

 

[paulmdrdo1]

Triangle $OAB$ is isosceles, so what must the other two angles be?

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

- - - Updated - - -

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?

 

[MarkFL]

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

$\overline{OA}$ and $\overline{OB}$ are both radii of the same circular segment, right?

 

[Fantini]

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 $BAC = 15^{\circ}$.