Points $ M,N$ are taken on sides $ BC,CD$ respectively of parallelogram $ ABCD$. Let $ E=BD\cap AM, F=BD\cap AN$. Diagonal $ BD$ cuts triangle $ AMN$ into two parts. Prove that these two parts have equal area if and only if the point $ K$ given by $ EK||AD, FK||AB$ lies on segment $ MN$.
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Tags: geometry, parallelogram, ratio, geometry proposed
hinhhoc273
18.06.2008 06:36
I've found a solution for this problem ( not very nice) First, we have AE/AM* AF/AN =1/2(1).BM=x, DN= y, AB= a, AD=b. From (1) we found that ab= ax+by+ xy.(2). Call I, K are points on MN such that EI//AD, FK//AB. We'll prove that I=K We call X, Y are intersection of MN with AD, AB. Then, we use Talet theorem to have: MI/MX= ME/MA= x/(b+x), but NX/NM= ND/NC= y/(a-y), so MI/ (MN+ MN(y/(a-y))) = x/b+x, then MI/MN = ax/(a-y)(b+x)= 1/2 (use (2)) Analogusly, we have NK/ MN = 1/2. So I=K. It's all the prove.
Sunkern_sunflora
18.06.2008 08:08
Quote: hinhhoc273 wrote: Then, we use Talet theorem to have: MI/MX= ME/MA= x/(b+x), but NX/NM= ND/NC= y/(a-y) What is Talet theorem???
hinhhoc273
18.06.2008 10:55
sorry everybody because I'm poor in English Talet theorem is use in Vietnamese. I's about the ratio of some parallel lines. Example: Let ABC be a triangle. a line call d is parallel with BC intersect AB, AC at M,N, then AM/MB= AN/NC, AM/AB=AN/AC= MN/BC
28121941
18.06.2008 11:05
In the western literatura, the name is Thales theorem. It was clear from the context of the solution.
Sunkern_sunflora
18.06.2008 18:09
Talet is a word of Vietnamese! Interesting! Will you send me some more such words with the actual english meanings, pleeeease!! Also, thanks for giving data about the theorem.