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This problem is extremely hard for us because we learned the third part far differently than you have so we have here a problem in communication. But we'll all try.

Circle Problem Diagram in the middle

The trick here is actually in reading the problem. You want the ratios always to be in the ratio of 1:2. That means smaller over larger and that must always be so.

A: A is not true. You are given 2 radii

B: B is not true. Or rather on 1 circumstance it could be true. That would be when BC = EF or GI. That would give you 1:2. But in general that is not so.

C: ...FA:AI both are radii. C is not true.

D: ...AI: EF The first is a radius. The second is a diameter. That is always 1 to 2 Answer.

E: AI:AD Two radii. The ratio is 1:1

Problem 2: Diagram left

These two triangles are drawn as shown in the diagram on the left. The green triangle has been rotated counterclockwise 90o. It has been rotated around the point (4,8). The triangles are congruent (good for you for getting it). I can't answer the second box because I don't know how you have been given the choices, but the way I have answered it is what has happened. The word rigid in this case means that the shape has not been altered.

Third problem. We have to use your diagram.

This is the problem that is going to give us the most trouble. I would solve this an entirely (and shorter) way.

AB || CD . . . . . . . . . . . . . . . Given

<AIE = <GJB . . . . . . . . . . . Given

1. <AIE = <IKC . . . . . . . . . .Corresponding angles of parallel lines are equal

2. <IKC = <JLD . . . . . . . . . Substitution for an equal by another equal and given

3. <ILK + <IKC = 180o . . . .Two angles on a line sharing a common side are supplementary. (Supplementary means they add to 180o)

4. <IKC = <JLD . . .. . . . . . Transitive property of equality.

5. <IKL + <JLD = 180 . . . . Step three and substitution of equals (transitive property of equality).

6. <IKL = <DLH . . . . . . . . .Supplements of equal angles are equal Transitivity.

just sum

10 + 10 = 20