
Join [XM], [MK], [KZ], and [ZX].
In △AOC, X and Z are midpoints of [AO] and [OC] respectively.
∴[XZ]∣∣[AC] and XZ =21AC {midpoint theorem}
Similarly, in △ABC,
[MK]∣∣[AC] and MK =21AC
Thus [XZ]∣∣[MK] and XZ = MK
∴XZKM is a parallelogram.
∴the diagonals [MZ] and [KX] bisect each other.
It can similarly be shown that LKYX is a parallelogram, and that [LY] bisects [KX].
∴[KX], [MZ], and [LY] all pass through the midpoint of [KX].
∴[KX], [MZ], and [LY] are concurrent.