Simple geometry question In the picture below you can see a triangle were $AB$ is equal to $AC$, so if $AOB$ is $x$ I understand why the other ones are too. But why is $OBC$ is $2x$?
Thanks.

 A: Look at the two skinny triangles. Their missing third angles are each $180-2x$, adding up to $360-4x$.
So the unwritten angle of the fattish triangle is $4x$. That leaves $180-4x$ to be split between the two bottom angle, $90-2x$ to each. Not $2x$. Unless, of course, $x=22.5$, which it needn't be.
Or else, better,  look at the big triangle. At the top it has angle $2x$. Instead of $2x$ for the bottom angles, which is wrong, write $y$ and $y$.
Then $2x+2(x+y)=180$, giving $y=90-2x$.
Remark: The first solution was a typical instance of angle-chasing. We filled in all the angles, finally getting to the ones of interest. Nothing wrong with angle-chasing, it is pleasantly algorithmic. In the second solution, we went directly for the goal. 
A: Without further information you don't know that it is.  As long as $B$ and $C$ are symmetric with respect to $OA$, the angles $OBC$ and $OCB$ are equal but can have a range of values.  Think of them first at $9$ o'clock and $3$ o'clock.  Then $x=45^\circ$ and $OBC=OCD=0^\circ$.  If they approach 6 o'clock, $x$ approaches $0^\circ$ and $OBC=OCB$ approaches $90^\circ$
