# angle of inscribed triangle

let us consider following problem:

we have inscribed triangle,like this

we are asked to determine if given $x$ is acute or obtuse,$O$ is center of circle,as i know if $CB$ would be diameter,then $x$ would be $90$,or $CAB$ would be right triangle,but in this case what it should be?if i trust diagram then maybe it seems that it is more then $90$,but because GRE said that dont trust diagram,then what i should know about it?

• @labbhattacharjee but how do you show that $CAB$ is greater than $DAE$? It seems like you are resorting to the diagram again. – BlackAdder Jul 28 '13 at 14:29

There is a proposition that says that if you connect $C$ to $O$ and $B$ to $O$, then the reflex $\angle BOC$ is twice of $\angle BAC$. And since reflex $\angle BOC >180^{\circ}$, we are done.
To argue that the reflex $\angle BOC >180^{\circ}$, we can just say that the usual $\angle BOC <180^{\circ}$, because it is contained in the triangle $BOC$.
• sorry $BOC$ is more then $180$? – dato datuashvili Jul 28 '13 at 14:39
• There are two angles when two lines meet at a point. Usually, what we call the "angle" is the inside angle, ie angle which is less than $180$. The reflex angle is the one which is greater than $180$. – BlackAdder Jul 28 '13 at 14:43
• you mean angle which is not in $BOC$ triangle? – dato datuashvili Jul 28 '13 at 14:45