Circle Geometry

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.

Could someone please help me? I can't seem to work this one out.

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)

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

Grandad