Assume a quadrilateral $ABCD$ and $M, N$ points on $AB$ and $CD$ respectively, such as $\frac{AM}{MB}=\frac{CN}{ND}$. Lines $AN$ and $MD$ intersect on $K$ and lines $MC$ and $BN$ intersect on $L$. Prove that the area $(KMLN)$ equals to the sum of the areas $(AKD) + (BLC)$.
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Have you checked the special case $M$ midpoint of $AB$, equiv. $N$ midpoint of $CD$? – hardmath Jan 21 '14 at 15:45
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yeap! have already checked all "good" cases, AM/MB=1 or 1/2 etc. I don't know if KMLN is a parallelogram though – jacie Jan 21 '14 at 15:48
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Assuming that KMLN is a //gm. Then, by intercept theorem, KN = AN/4. Similarly, ML = MC/4. By properties of //gm, KN = ML. This further means AN = MC which in general is not necessary true. Thus, KMLN is not a //gm in general. – Mick Jan 21 '14 at 16:58
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The '4' should be r + 1; where r = AM : MB. Sorry – Mick Jan 21 '14 at 17:46
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denote $\triangle ABC=$ area of triangle $ABC$
$\triangle AKD=\triangle ADN-\triangle KDN, \triangle BLC=\triangle BMC-\triangle BML$
$\triangle KLN=\triangle MDN-\triangle KDN,\triangle MLN=\triangle BNM-\triangle BML$
$\triangle AKD+\triangle BLC=\triangle KLN+\triangle MLN \iff \triangle ADN+\triangle BMC=\triangle MDN+\triangle BNM \iff \dfrac{DN}{NC}\triangle ANC +\dfrac{BM}{AM}\triangle CMA=\dfrac{DN}{NC}\triangle MNC+\dfrac{BM}{AM}\triangle NMA $
$\dfrac{DN}{NC}=\dfrac{BM}{AM}=p$
LHS$=p\triangle ANC+p\triangle CMA$, RHS$=p\triangle MNC+p\triangle NMA$
$\iff \triangle ANC+\triangle CMA=\triangle MNC+\triangle NMA$
the last one is true.
QED
chenbai
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You mean $\triangle AKD = \triangle ADN -\triangle KDN,~ \triangle BLC = \triangle BNC -\triangle LNC, \ \triangle MKN = \triangle MDN -\triangle KDN,~ \triangle MLN = \triangle MNC -\triangle LNC$, right? – jacie Jan 23 '14 at 11:01
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I also cannot understand how $\triangle BNC =\frac{BM}{AM} \triangle CMA$ and $\triangle MNC =\frac{BM}{AM} \triangle NMA$, since the altitude from $B$ to $CD$ is not equal to the one from $C$ to $AB$. – jacie Jan 23 '14 at 11:14
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