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
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 =/
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)