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

• Jul 7th 2010, 10:06 PM
Lukybear
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.
• Jul 7th 2010, 10:17 PM
Also sprach Zarathustra
Hint:

• Jul 7th 2010, 11:30 PM
Lukybear
Hmm... my X is outside the circle so it definately does not equate to radius. Do you have any other ideas?
• Jul 8th 2010, 03:26 AM
sa-ri-ga-ma
AB is parallel to CD.

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

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.
• Jul 8th 2010, 05:45 AM
Lukybear
So the reasnoning here is basically that; since YC = YD, this can only at one point in the circle. Therefore Y is O?
• Jul 8th 2010, 08:07 AM
sa-ri-ga-ma
Quote:

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.
• Jul 8th 2010, 05:39 PM
bjhopper