geometry problem?

Question

HI,

I am not able to solve this geometry problem.

http://img149.imageshack.us/my.php?image=scan00016ti.jpg

i can see AOC ,BOC,AOB are isoceles triangle because they are having same radius as their 2 sides.


but my book says, ABC is also isoceles apart from the above triangles. can you explain why ABC is isoceles ?



here is my book's solution which i dont understand.

combining I and II ,


so,ABC is also isoceles.

Answer 1

it's given
a theorem says that

a chord makes an angle at the center is twice that makes an angle in the remaining part of the circle
(please verify the correct sentence)
ie

there fore
now



now the triangle OCB is isoceles
and hence OC= OB


the total angle of a triangle is 180 degree
and hence Similarly there fore ABC is an Isoceles triangle

Answer 2

ok, this might be confusing, but i hope you can understand! I don't know if it'll help, and it might not be perfectly right, but i still got an issoceles triangle.

Ok,
ok, i hoped that helped, and i hope it wasn't too confusing. It really helps to draw it out, since math is kind of a visual process. I hope you figure it out!

Answer 3

To find the number of isoceles triangles, you first need to determine which legs in the figure are the same length.

1) Lines OA, OB, and OC are the same length because they are all radii of the same circle. Therefore, the three interior triangles that have O at the vertex are all isoceles.

So, the trick is to see if triangle ABC is also isoceles.

2) Focus on the triangle OCB for the moment. Since OC=OB, that triangle is isoceles. Knowing that, angle
Now find angle BOC. Since the sum of the interior angles of a triangle = 180 degrees, 180 =
3) Now, lets figure out what angle
Solving for
4) Now, show that at least two of the legs of triangle ABC are equal.

Consider triangles AOC and BOC. We know that angles
Hope this helps,

-Guru

Answer 4

Here is a relatively simple proof of their assertion:

Notice that angles ACB and AOB both subtend the same arc, AB. ACB is an inscribed angle and AOB is a central angle. It can be proven that the measure of an inscribed angle is half the measure of its corresponding central angle. Since it is given that the measure of central angle AOB is 60­°, that forces the measure of its corresponding inscribed angle, ACB, to be 60­°/ 2 = 30°. The measure of angle OCB is 15­°. So the measure of angle OCA must also be 15° (30 - 15 = 15). Since triangle AOC is isosceles, and angles OAC and OCA are base angles of that triangle, their measures are equal. Therefore angle OAC = angle OCA = 15°. Now, because triangle BOC is also isosceles, for the same reason as triangle AOC (their legs are all equal to the radius of the circle), and one of its base angles, OCB = 15°, then its other base angle, OBC, must be 15°. Also, since you realize triangle AOB is isosceles, its base angles are also equal. So, we have added the same angle (15°) to the base angles of an isosceles triangle, which are also equal. Mathematically speaking:

m /_BAC = m /_BAO + m /_OAC = m /_ABO + m /_OBC = m /_ABC.

Therefore, by transitivity: m /_BAC = m /_ABC.

In other words, those angles, which are the base angles of another triangle, are equal, forcing that triangle to also be isosceles. So triangle ABC is an isosceles triangle.

Answer 5

CO will bisect the chord AB perpendicularly
if CO meets AB at D AD=DB
and CD is common
so the triangles CAD and CBD are congruent
So CA=CB (c.p.c.t)
so triangle ABC isosceles