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Circle Geometry (1 Viewer)

TrueHappiness

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AB and AC are equal chords of a circle. AD and BE are parallel chords through A and B respectively. Prove that AE is parallel to CD
 

MetroMattums

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Construct lines AE, CD, BC (extend BC to X)

Let the intersection of AE and BC be F, < ABC = a, < CBE = b

So < ABC = < ACB = a (equal <'s opp equal sides)
But < ACB = < AEB = a (<'s subtended by same arc)

Considering ∆ BEF, < BFE = 180 - (a + b) (< sum of ∆)
So < AFC = 180 - (a + b) (vert opp <'s)

But < BAD = 180 - (a + b) (co-int <'s, AD||BE)

So < BAD = < DCX = 180 - (a + b) (ext < of cyclic quad ABCD)

ie. < AFC = < DCX

ie. AE||DC (corresp <'s)

If I had a diagram that would make it much easier :|
 
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