Thread: Circle Geometry: Proving a line passes through centre of circle

AB and CD are two parallel chords of a circle. CA and DB meet, when produced, at X. AD and BC intersect at Y

There are three parts to this question, two of which are already solved and i will post results:

a)YA= YB
b)XA= XB

How must i prove that XY will pass through center?

And just generally, how is this achieved without using co-ordinate geometry, i.e. proving a certain line passes through centre.

Hint:

prove XY radius .

Hmm... my X is outside the circle so it definately does not equate to radius. Do you have any other ideas?

AB is parallel to CD.

So

ABDC is cyclic quadrilateral.

So

Hence

on the chord AB are equal.

Hence . So yC = yD.

If O is the center of the circle, then OC = OD. It is possible only if O and y lie on the perpendicular bisector of CD.

Hence xy passes from the center of the circle.

So the reasnoning here is basically that; since YC = YD, this can only at one point in the circle. Therefore Y is O?

Originally Posted by Lukybear

So the reasnoning here is basically that; since YC = YD, this can only at one point in the circle. Therefore Y is O?

Y is not O. Here OC = OD = radius of the circle. And YC = YD. It is possible when Y lies on OX.

[ a wordy answer- interested in comments

two parallel chords subtend equal arcs therefore ABCD is a regular trapezoid. Its diagonals are equal and meetalong the perpendicular bisector of AB,CD and the center of circle.
angle ACD = angle BCD(equal arcs) triangle XCD is isosocles therefore X lies along the same perpendicular bisector


bjh