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axwe7

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

BLIT2014

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In Triangle ACE:

Angle E = 90°(given)

=>Angle ACD+Angle BAC = 90° (Angle sum of a triangle)

Angle AOD = 2 Angle ACD (The angle at centre must be equal to twice the angle at circumference)

So Angle BOC = 2 Angle BAC (The angle at centre must be equal to twice the angle at circumference)

Angle AOD + Angle BOC = 180°
 

Drongoski

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Say you have the diagram, appropriately labelled. (I have chord AB horizontal, below O, and vertical chord CD, to the right of O, with C on top) Draw the line AC.

Then: angle AOD = 2 x angle ACD = 2 x @ say

and angle BOC = 2 x angle BAC = 2 x & say.

Now in right-angled triangle ACE: @ + & = 90 deg

.: angle AOD + angle BOC = 2 x (@ + &) = 2 x 90 deg = 180 deg
 
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