# Circle Geometry

• May 19th 2009, 12:54 AM
xwrathbringerx
Circle Geometry
Hey guys

Two perpendicular chords AB and CD of a circle, centre O, intersect at E. Prove that $\angle AOD$ + $\angle BOC$ = 180 degrees.

• May 19th 2009, 04:08 AM
Hello xwrathbringerx
Quote:

Originally Posted by xwrathbringerx
Hey guys

Two perpendicular chords AB and CD of a circle, centre O, intersect at E. Prove that $\angle AOD$ + $\angle BOC$ = 180 degrees.

In $\triangle ACE$:

$\angle E = 90^o$ (given)

$\Rightarrow \angle ACD+\angle BAC = 90^o$ (angle sum of triangle)

But $\angle AOD = 2\angle ACD$ (angle at centre = twice angle at circumference)

and $\angle BOC = 2\angle BAC$ (angle at centre = twice angle at circumference)

$\Rightarrow \angle AOD + \angle BOC = 180^o$