you're referring to quad. HGCD right? so if H and G is cyclic for a pair of points in the circumference, then it means that H and G is always cyclic with any other points in the circumference?All you need is a quadrilateral with a pair of opposite angles supplementary .
O_O how do u know BAD + HGB = 180Nope, no reference to HGCD.
BAD + HGB = 180, so ABGH has a pair of opposite angles which are supplementary. This is one characterisation of cyclic quads.
Not sure what you mean by "H and G is cyclic for a pair of points in the circumference"...
O_O alright thanks, how do i go about proving them lolThat the two points are the same.
oh alright thanks! ill think a bit more about it 2morroWell if cd and dx are parallel and they both intersect at D then they must be the same line.
ohhh alrite thxProve angle rap is 180 degrees? That means that points r, a and p are all on the same line since 180 degrees is a straight line. Therefore they are collinear