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.
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