Thread: Another Geometry Proof (Circles and Chords)

So the same situation as the last..but new proof:

Given: Tangent line AE at B. Secant EC, which meets the circle at D.

Prove: <ABC = <BDC and <EBD = <BCD

Originally Posted by Volux

Given: Tangent line AE at B. Secant EC, which meets the circle at D.
Prove: <ABC = <BDC and <EBD = <BCD

I will give you three basic facts. You use them to do the rest.

By using the 2m(<BED) and etc..do you mean 2 of angle BED?

And wouldn't m(arc(CB)) - m(arc(BD)) essentially mean just arc CD?

Originally Posted by Volux

By using the 2m(<BED) and etc..do you mean 2 of angle BED?
And wouldn't m(arc(CB)) - m(arc(BD)) essentially mean just arc CD?

The m stands for measure.

Are you sure that you know enough geometry to tackle these problems?
You seem to be short on the basics.

Yes I know what m means, though we aren't required to use them in class. We have never actually used 2 angles for one thing either.

I know geometry somewhat, it just happens to be that I have an A- in the class and got the harder proofs in which I am having a hard time with.

I'm not too great at where to start and justifications..or proofs in overall. Not great at them =/

I know arc BD = <BDC. and arc CD = <CBD.

How am I to put those together though?

Would this work for the proof?:

1. given stuff..
2. <EBD = (1/2)(arc)BD 2. theorem used in the last proof I gave
3. <BDC = (1/2)(arc)BC 3. I AM UNSURE..help??
<DBC=(1/2)(arc)DC
<BCD=(1/2)(arc)BD
4. <ABC=(1/2)(arc)BC 4. theorem used in the last proof i gave
5. <ABC=<BDC 5. substitution
<EBD=<BCD


Also can you help me out with this next proof? SAME image as before and same given, except:

PRove: BE(SQUARED) = EC (TIMES) ED

Hint: Use similar triangles, and the theorem from #2(the proof above)

Hi volux,
Which triangles are similar.Define them and use proportions