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

1. ## 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.

2. Hint:

prove XY radius .

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

4. AB is parallel to CD.

So $\displaystyle \angle{BAD} + \angle{DCA} = 180^o$

ABDC is cyclic quadrilateral.

So $\displaystyle \angle{BAC} + \angle{BDC} = 180^o$

Hence $\displaystyle \angle{ACD} = \angle{BDC}$

$\displaystyle \angle{ACB} and \angle{ADB}$ on the chord AB are equal.

Hence $\displaystyle \angle{yCD} = \angle{yDC}$. 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.

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

6. 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.

7. [ 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

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