Proving algebraic equations with circle theorems 
I got as far as stating that OBP=90˚ (as angle between tangent and radius is always 90˚), and thus CBO=90˚- 2x. CBO=OCB as they are bases in a isosceles. COB=180-90-2x-90-2x. But after this, i am clueless.
I am stuck with this Question. It is from a GCSE Further Maths past paper. Despite seeing online tutorials, and checking the answer scheme, I still don't understand how you solve this question. Could you please show me a step by step explanation of how you solve this question. Thank you.


ANSWER:

 A: At first forget about the condition that $\angle CBP = 2x$. (The condition $\angle CBP = 2 \angle ODC$ determines the exact position of C.)
You can express $\angle BCD$ with $y$. Using $\angle OCD = \angle ODC = x$ you can express $\angle OCB = OBC$ with $x$ and $y$ and hence also $\angle CBP$. 
Now use the condition that $\angle CBP$ equals $2x$. This gives an equation in $x$ and $y$.
A: Try expressing the same angle both in terms of $x$ and in terms of $y$. Then you can set the expressions equal to each other and rearrange. For example, you can do this with either $\angle BOD$ (either the reflex or obtuse angle) or $\angle BCD$. Does this help?
A: Angle BOD = 180 − y
Angle OCD = x
Angle OBC = 90 − 2x
Angle BCO= 90 − 2x
Angle BOD reflex = 360 - (90 − 2x) − (90 − 2x) − x − x = 180 + 2x.
180 − y + 180 + 2x = 360, 
thus  y = 2x
A: I am also doing Further Maths GCSE and what i would do is draw a line from B to D (this is a lot shorter) this would create the alternate segment theorem for angles therefore $BDC$ would be $2x$ and $BDO$ and $DBO = x$ which means $ BOD = 180- 2x$ which means $y = 360 − 90 − 90 − (180 − 2x)$
$y = 2x$
