Given a circle and two points on that circle,

*P* and

*Q*, draw the chord

*PQ*, and label its midpoint

*M*.

Now draw two other chords of the circle *AB* and *CD* that both pass through *M*.

Further, draw chords *AD* and *BC*.

Label the intersection of *AD* and *PQ*, point *X*.

Label the intersection of *BC* and *PQ*, point *Y*.

_____________________________

Prove that *M* is the midpoint of line segment *XY*.

well... there is a sure-fire way to get MX = MY, which is CPCTC (correspoding parts of congruent triangles are congruent) after either figuring out if triangle CMY is congruent to DMX, or figure out YMB is congruent to XMA... hmm i already notice vertical angles, so mebe u gotta use SAS or SAA?

also... u no that angle D is congruent to angle B because they share the same arc CA. Similarly, angle C and A are congruent because they share the same arc BD.

one more thing... u can prove that triangle CMB is similar to AMD by AAA, but their congruence cant be proven just yet (if possible)

*Edited on ***March 24, 2004, 4:42 pm**