The opposite angles in a cyclic quadrilateral add up to [math]180^\circ[/math][br][br]In plain terms:[br][br]If you look at the cyclic quadrilateral (a four sided shape where the four vertices touch the circumference of a circle) below, each vertex is a point on the circumference of the circle. I have divided the quadrilateral down the middle with a dotted line (take note that the line does [i]not [/i]pass through the centre of the circle so it does [i]not [/i]represent the diameter of the circle). The dotted line divides the quadrilateral into two triangles. Remember, all the internal angles of a triangle add up to [math]180^\circ[/math]. Therefore, all the internal angles of a cyclic quadrilateral will add up to [math]360^\circ[/math][br][br]If you add Angle [math]\alpha[/math] and Angle [math]\beta[/math] the result will always be equal to [math]180^\circ[/math][br][br][math]\angle\alpha[/math] + [math]\angle\beta[/math] = [math]180^\circ[/math][br][br]If you add up the angles for the remaining two unknown angles then that will also equal [math]180^\circ[/math][br][br]Scroll [i]all the way [/i]down for more information!

What else is there to say? I'll tell you... Take a look at the graphic below and note the relationship between Angle [math]\alpha[/math] and Angle [math]\beta[/math] ( [math]\angle\alpha[/math] and [math]\angle\beta[/math] )and the relationship between Angle [math]\gamma[/math] and Angle [math]\delta[/math] ( [math]\angle\gamma[/math] and [math]\angle\delta[/math] )[br][br][math]\angle\alpha[/math] = [math]\angle\beta[/math] and [math]\angle\gamma[/math] = [math]\angle\delta[/math][br][br]Does this remind you of anything? It should because if you look back at [url=https://www.geogebra.org/m/nx5PMCzT#material/wCGeV5ku]Circle Theorem 2[/url] then you will see the relationship between two adjacent vertices, the arc between them and the other two vertices on the opposite side of the circle.