cyclic quad question (1 Viewer)

[5647382910]

say you have a quadrilateral ABCD in order (clockwise). If angle DBC = angle CAD can we then say that ABCD is cyclic as if you draw a circle, angles in the same segment are equal?
why/why not?
thanks in advance

 

[Rachaek]

Yep, I'm pretty sure it is a cyclic quadrilateral. As you say, angles in the same segment are equal. Since this is a property shown by points lying on the circumference of a circle, it can be said that these points are concyclic (ie. form a cyclic quadrilateral)

also, it says on wikipedia, therefore it must be true
check out the picture at the bottom http://en.wikipedia.org/wiki/Concyclic_points

 

[Michaelmoo]

say you have a quadrilateral ABCD in order (clockwise). If angle DBC = angle CAD can we then say that ABCD is cyclic as if you draw a circle, angles in the same segment are equal?
why/why not?
thanks in advance

Yep, Im 100% sure you can do this, Basically there are 3 ways to prove a quadrilateral is a cyclic quad:

i) Angles in the same segment standing off the same arc are equal, therefore points of arc and subtended to form a cyclic quad (your situation)

ii) Opposite angles of a cyclic quad are supplementary, i.e. in your case < DAC + < BCD = 180 degrees

ii) Exterior angle of a cyclic quad is equal to the opposite interior angle. i.e. if you produce DC to X in yuour situation <BCX = < BAD

Hope that helps, Good luck.