定理　凸四边形ABCD的两条对角线交于BD的中点O ,过O的两直线分别交BC于E、AD于F ,交AB于G ,CD于H ;GE、FH分别交BD于M、N ,则MO =ON .证明 :如图 ,过E、F、G、H作AC的平行线 ,分别交BD于E′、F′、G′、H′,再引BD的垂线 ,垂足依次为E1、F1、G1、H1.由于 OM(EE1+GG1)ON(FF1+HH The two diagonals of the theorem convex quadrilateral ABCD intersect at the midpoint BD of the BD. The two lines passing over O cross the BC at E and AD at F, the AB at G, and the CD at H; the GE and FH intersect with the BD at M respectively. N, then MO = ON. Proof: As shown in the figure, E, F, G, and H are the parallel lines of the AC, and the BDs are respectively transmitted to E′, F′, G′, H′, and then the vertical line of the BD. The order of feet is E1, F1, G1, H1. Since OM (EE1 + GG1) ON (FF1 + HH)