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The parallelogram ABCD, e, f are the trisection points of BC, connecting BD, AE, AF to intersect BD with m, N, finding: Mn: Nd It is known that: the parallelogram ABCD, f is the trisection point of BC, connecting BD, AE, AF with BD in M, and finding: Mn: Nd
In parallelogram, ad ‖ BC ad = BC
∵ e, f are the trisections of BC
∴BE=1/3BC=1/3AD BF=2/3BC=2/3AD
∵AD∥BC
∴DN/BN=DA/BF=3/2,DM/BM=DA/BE=3
∴DN=3/5BD BM=1/4BD
∴MN=BD-DN-BM=3/20BD
∴MN∶ND=3/20∶3/5=1∶4
In square ABCD, a point E on BC, a point F on CD, the angle EAF = 45 degrees, connect BD, intersection AE, AF with m, N, find the relationship between BM, Mn, nd
1:1:1
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