prove XY radius .
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:
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.
AB is parallel to CD.
ABDC is cyclic quadrilateral.
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.
[ 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