Simple Circle Geometry Question (1 Viewer)

[bawd]

The quadrilateral ABCD is a cyclic parallelogram. Prove that it is a rectangle.

Easy enough question, but my brain is currently fried. I know it's something to do with the external angle of a cyclic quadrilateral, followed by supplementary angles of a cyclic quadrilateral and then co-interior angles on parallel lines, but it seems wrong no matter how I word it.

 

[AkaiHanabi]

the opposite angles of a parallelogram are always the same. also, the opposite angles in a cyclic quad = 180

therefore if one angle is x, the opposite angle is also x

x + x = 180

therefore x = 90

similarly for the adjacent angle, since it is a parallelogram, it is cointerior. lets say the angle is y

so x + y = 180, but we know x = 90, therefore y=90

since all angles are 90 degrees, and it is a parallelogram, it is a rectangle (all angles =90, opposite sides are equal, etc)

 

[bawd]

the opposite angles of a parallelogram are always the same. also, the opposite angles in a cyclic quad = 180

therefore if one angle is x, the opposite angle is also x

x + x = 180

therefore x = 90

similarly for the adjacent angle, since it is a parallelogram, it is cointerior. lets say the angle is y

so x + y = 180, but we know x = 90, therefore y=90

since all angles are 90 degrees, and it is a parallelogram, it is a rectangle (all angles =90, opposite sides are equal, etc)

Thanks! That's a much easier way of proving than what I was thinking of using. (forgot to look at all the properties of a parallelogram)

 

[AkaiHanabi]

Thanks! That's a much easier way of proving than what I was thinking of using. (forgot to look at all the properties of a parallelogram)

No problem =) Have fun with geometry, I sure didn't *shudders* XD