Thread: [SOLVED] help on proof!

i need to prove this as many ways possible, and i need some new approaches...

BC is the diameter of a circle with a center O.
A is a point on the circle.
OE and OD bisect chords AB and AC.
Prove OE is perpendicular to OD.

you'd think analytical would work the best but im having trouble.

Here is one. It is not in your "two-column statement-reason" style.
See if you can supply the "reasons" anyway.

Imagine, or draw the figure on paper.
It is a circle BAC whose center is O.
Triangle BAC is inscribed in the cirle.
BOC is a diameter.
In side BA, point E is in the middle, so BE=EA.
In side CA, point D is in the middle, so CD=DA.

Draw radius OA.

In sector BOA,
Chord BEA is bisected by point E.
Draw radius OEF, where point F is on the arc BA.
Since the radius OEF bisects the chord BEC, then this same radius OEF bisects arc BFA.

In sector COA,
Chord CDA is bisected by point D.
Draw radius ODG, where point G is on the arc CA.
Since the radius ODG bisects the chord CDA, then this same radius ODG bisects arc CGA.

In sector FOG,
Central angle FOG is subtended by arc .
If we can show that this arc measures 90 degrees, then it means central angle FOG is 90 degrees too.

IN semicircle BFAGC,
Arc BFAGC = semicircle = 360/2 = 180 degrees
arc BF = arc FA
arc GC = arc AG
So, arc BFAGC = arc BF +arc FA +arc AG +arc GC = 180 deg
Substitutions,
arc FA +arc FA +arc AG +arc AG = 180 deg
2(arc FA) +2(arc AG) = 180 deg
2(arc FA +arc AG) = 180 deg
(arc FA +arc AG) = 180/2 = 90 deg
So,
arc = 90 deg.

So, central angle FOG is 90 degrees

That means angle EOD is 90 degrees also.

That means OE is perpendicular to OD.

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I hate Proofs. I hate memorizing definitions.
I like calculations or computations.