In this pic, we are arc BC> arc CD, indicating that angle BOC > angle COD. Along this vein. we should well know that since angle BOC > angle COD, then x would be <y, since these are insribeded isolosceles tiangles.

Bump for further discussion.

Regards

We can take the edge case for BC > CD. Let C be very close to D. Then y goes to 90 degrees whereas x becomes very small. Therefore, x has to be smaller than y always since they'll only be equal if BC = CD (as OB = OD)

I think that a good way to see it is with the concept of what an angle is:

The lower the angle that a chord forms with the diameter, the longer the chord, being a 0º angle the diameter itself. So, if BC>CD, it must mean that x<y as both BO and DO are radii, hence answer B

I do not quite understand. Isn't OC the radius, and the angles x and y are opposite to the radius, and so shouldn't x=y, the length of BC and CD shouldn't matter right? Please explain why my logic is wrong.

The lengths of BC and CD matter the most

Hint: The angle opposite the longest side is always the biggest angle.


For more on this hint, watch the following (starting at: 1:00)



Cheers,
Brent
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Angle BOC is greater than angle COD. So 180-2x> 180-2y, which refers to x<y. So, B is greater

Angles in the same segment are equal. Shouldn't C be the answers since x & y are in same segment ?

No because \(BC > CD\)

I did not do this problem the best way. I saw this problem as many other similar geometry problems as what I call "Stretch and Squash". Visually it is difficult to see and therefore I make a new figure where I take the dimensions to the extreme. Draw a right triangle and then draw another one with the leg extended way out. You can see the angle gets smaller as a result. Therefore, we can also say that angle BCO is smaller than angle CDO and then again use the isosceles triangle point made earlier. Again, this problem is better done as mentioned, but the concept of stretch and squash can be useful.