Chord AB is inscribed in a circle. It is bisected. Leg OB is the hypotenuse of a triangle. leg OB and leg OX are equal. So an isosceles triangle. no values are given. Leg O is the perpendicular bisector of chord AB. Can I find the measure of arc AB with only the value of the 90 degree angle of the the bisector? if so, how?
Hi West Chicago, IL...Itasca here.
I'm having the same trouble as the others are in figuring out what the drawing looks like. Here's what it sounds like:
- There's a chord drawn in the circle. That's AB.
- The center of the circle is O.
- The bisector of AB is drawn. And you pointed out a conclusion you can make from that, which is that the bisector is perpendicular to AB, so it forms a right triangle, with OB as the hypotenuse.
- You said there's an isosceles triangle somewhere. But that's where I get confused.
So, where's point X, and where are the two legs of the isosceles triangle? I'll take a guess...that X is where the bisector intersects AB. So, you have a right triangle, with OX (along the bisector) and BX (along the chord) being the legs, and OB (from the center to one end of the chord) is a hypotenuse.
If that's true, then:
- draw in the segment OA (from the center to the other end of the chord). Now, you've got two triangles, OXA and OXB.
- I keep saying that there's a segment OB, from the center to B, and then OA, from the center to A. What's a segment called that is drawn from the center of the circle to the edge of the circle? And what do you know about those segments?
- What do you know about those two triangles? What angles do you know, and are there any segments in your diagram that are congruent?
- Based on that, it'd be nice if you could figure out the angle formed by AOB...the central angle that creates arc AB. Because, if you know the central angle, then doesn't that tell you the angle of the arc that you're looking for? Central angles are equal to the arc they create...
Hope this helps,