Hey guys

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

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

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- May 18th 2009, 11:54 PMxwrathbringerxCircle Geometry
Hey guys

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

Could someone please help me? I can't seem to work this one out. - May 19th 2009, 03:08 AMGrandad
Hello xwrathbringerxIn $\displaystyle \triangle ACE$:

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

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

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

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

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

Grandad