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So many midpoints!

Source: 1983 National High School Mathematics League, Exam Two, Problem 3

February 22, 2020
geometry

Problem Statement

In quadrilateral ABCDABCD, S△ABD:S△BCD:S△ABC=3:4:1S_{\triangle ABD}:S_{\triangle BCD}:S_{\triangle ABC}=3:4:1. M∈AC,N∈CDM\in AC,N\in CD, satisfying that AMAC=CNCD\frac{AM}{AC}=\frac{CN}{CD}. If B,M,NB,M,N are collinear, prove that M,NM,N are mid points of AC,CDAC,CD.
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