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Given any convex quadrilateral $ABCD$ and an inner point $z$ is it possible to draw/construct a line $EF$ passing through $z$ s.t. $$\frac{AE}{EB}=\frac{DF}{FC}$$

quadrilateral

user149901
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4 Answers4

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The following might work:

Prove this for a square (trivial) Then use the result that any convex non-degenerate quadrilateral can be obtained by a perspectival (projective) transform from the square and that the ratios will be preserved.

Conrad Turner
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  • Actually I came to this problem when trying to prove that that projective map is surjective :) How to prove that $z$ has a pre-image in $[0,1]^2$? Please have a look to Surjectivity – user149901 Dec 21 '14 at 05:38
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I don’t think it is always possible.

enter image description here

Let the red line (through Z) be that required line. It cuts AB and DC at E and F respectively.

Through E draw EG // AD cutting DC at G. It also cuts BD at E’

According to the requirement,

$\frac {DF}{FC} = \frac {AE}{EB} = \frac {DE’}{E’B}$ … [intercept theorem]

Through B, draw BX // E’F cutting DC produced at X.

BY similar triangles, $\frac {DE’}{E’B} = \frac {DF}{FC + CX}$

In other words, FC is too short for the equality to hold unless C is actually X.

Mick
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  • $\frac {DF}{FC} = \frac {DE’}{E’B} \implies E'F//BC$ – chenbai Dec 22 '14 at 04:11
  • @chenbai Don’t quite understand your comment. I try to explain my over-simplified description (sorry) first. According to the requirement, we must have $\frac { DF } { FC } = \frac {AE}{EB}$. On the other hand, by intercept theorem, we found $\frac {AE}{EB} = \frac {DE’}{E’B}$. Thus, if the red line works, we must have $\frac {DF}{FC} = \frac {DE’}{E’B}$. For the last statement to be true, according to the construction made, C must be extended to X. – Mick Dec 22 '14 at 06:34
  • Thanks for the reply, but I don't think this proves that proportionally cutting line doesn't always exist. This simply proves that X and C should coincide. If you change the draw of EF by moving E up a liitle bit and F down we get the right pcture: E′F//BC and there is no contradiction anymore... – user149901 Dec 23 '14 at 00:12
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EDIT1: If we only want to prove if $ EF$ exist,that $z$ is on $EF$, it is easy.

let $A(0,0),D(x_1,0),C(x_2,y_2),B(x_3,y_3),Z(a,b),H(p,q),H$ is on $AC$with $FH//AD \to EH//BC$, if we can find $H(p,q)$, then we can draw $EF$ and the problem is solved.

with the conditions, we have:

$((x_2-x_1)y_3-x_3y_2)p^2+((x_1-a)x_2y_3+ax_2y_2+bx_2x_3-bx_2^2+bx_1x_2)p-bx_1x_2^2=0 \\ q=\dfrac{y_2p}{x_2}$

which means $H$ is constructible even by compass and straight $\implies EF$ is constructable also.

The $p$ has two solutions because we didn't limit $E,F$ 's position and another solution is for $E,F$ out of segments$AB ,CD$ but on lines $AB ,CD$

chenbai
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  • @user149901,I didn't explain clearly and had edit it. – chenbai Dec 23 '14 at 05:22
  • Thanks for the answer. Why does the quadratic equation always have a solution? Can we prove that the discriminant is always $\ge 0$? – user149901 Dec 23 '14 at 07:39
  • Yes, for sure. But it looks ugly and I will try to find a nice method to show it. – chenbai Dec 23 '14 at 08:23
  • Would be very nice... because I stuck on that part. – user149901 Dec 23 '14 at 08:25
  • $\Delta=(x_1y_3-bx_3+bx_2-bx_1)^2+a^2(y_3-y_2)^2-2(y_3-y_2)(ax_1y_3-2bx_1x_3+abx_3-abx_2+abx_1)>0 \iff |a(x_1y_3-bx_3+bx_2-bx_1)(y_3-y_2)|> (y_3-y_2)(ax_1y_3-2bx_1x_3+abx_3-abx_2+abx_1)$ – chenbai Dec 23 '14 at 13:10
  • So it seems that even when $ABCD$ is not convex such a line always exists. Only this time it does not necessarily completely lies in $ABCD$, right? – user149901 Dec 23 '14 at 22:31
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Start with $E=A$ and $F=D$, so that $AE/EB=FD/FC$ because both are equal to $0$, and then move $E$ and $F$ toward $B$ and $C$ respectively, preserving the equality of the ratios (which go to infinity as $E\to B$ and $F\to C$). By continuity, the line $EF$ sweeps out the entire convex quadrilateral, hence at some point passes through the point $z$.

Barry Cipra
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