In a circumference with center O we draw two equal chord AB\equal{}CD and if AB \cap CD \equal{}L then AL>BL and DL>CL
We consider M∈AL and N∈DL such that \widehat {ALC} \equal{}2 \widehat {MON}
Prove that the chord determined by extending MN has the same as length as both AB and CD geometrytrapezoidangle bisectorgeometry unsolved