Semicircle problem

[Maratha1]

Hello Can anyone please help on this?

[BuckeyeT]

The initial question tells us quite a bit about the segment lengths. If O is the center of the semicircle with B, C, and D as points on the semicircle. Each segment OB=OC=OD (as they are each a radius of the circle). The question also states that AB=OC. So, AB=OC=OB=OD.

If you separate the image into triangles, you can see that you have three isosceles triangles (COD, BOC, and ABO).

In COD, OC=OD and LODC=LOCD (property of isosceles).
In BOC, OB=OC and LOBC=LOCB.
In ABO, AB=OB and LBAO=LBOA.

(1) LCOD = 60
Therefore, LODC = 60 and LOCD = 60 (since LODC=LOCD and all three must total 180).

Unfortunately, this only tells us that the other angles LCOB and LBOA together equal 120 (since LCOB + LBOA + LCOD = 180). They could equal 60 each, but they could equal 80 and 40. Insufficient.

(2) LBCO = 40
Therefore, LCBO = 40 and LBOC = 100. Again, this doesn't tell us anything specific about the other angles from O other than LCOD and LBOA together equal 80 (since LCOB + LBOA + LCOD = 180). Insufficient.

(1) and (2)
With both combined, we realize that LBOA = 180 - 100 (LBOC) - 60 (LCOD) = 20.

Since Triangle ABO is isosceles (with AB=OB), we know that LBOA = LBAO. LBOA = 20 = LBAO. Sufficient.

C.

[bourne3]

This has been discussed before on various threads.
One of them is here https://www.beatthegmat.com/question-fro ... 24724.html

This forum has great search capability. You can use that before you post the question

[BuckeyeT]

Ok, that one hurts my head. I see where I messed up. I forgot in my diagramming that segment AB continued straight through to form AC.

Therefore, it allows us to make the statements relating the equal angles of ABO and BOC.

The explanation in your link is pretty "light". I still had to work through a bit to see what was done. So, it's by no means a great example to learn from.