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ABC is a triangle and XY is variable straight line parallel to AC meeting BC and BA in X, Y respectively. If AX and CY meet at P find the locus of P.

Apoorv Jain
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3 Answers3

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Construction: Through P, draw MN parallel to AC meeting BC, BA at M and N respectively.

enter image description here

For some reasons, ⊿YNP ~⊿YAC. Therefore, $NP = AC. \frac {YN}{AY}$.

Similarly, we have $PM = AC. \frac {XM}{XC}$.

By intercept theorem, $\frac {YN}{AY} = \frac {XM}{XC}$.

∴ $NP = PM$ implying that P is the midpoint of the side NM of ⊿BNM.

Different sets of X,Y (i.e. X’, Y’ say) will generate the corresponding P’ of N’M’ of ⊿BN’M’.

∴ The locus of P is the median from B to AC.

Mick
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First, I assume you want the intersection between $CY$ and $AX$, because $XY$ and $CY$ intersect at $Y$.

The answer in this case is the "mediane"-line from $B$ to the middle point of $AC$.

Proof. The construction is invariant under affine transformations:

Given $A,B,C$ not colinear, the parallel $XY$ to the edge $AC$, is determined by the ratio of the lenghts $AY/AB$.

Given any other three points $a,b,c$ there is a unique affine map $f$ so that $f(a)=A,f(b)=B,f(c)=C$. Points $X,Y$ corresponds to $X=f(x)$ and $Y=f(y)$ and the proportions are preserved: $AY/AB=ay/ab$ and $CX/CB=cx/cb$.

So, you can solve the problem with an equilateral triangle, for which you see immedately that the locus of the $p$'s is the $b$-median line. Since the $b$-median line is defined as the line that joins $b$ with the midpoint of $ac$, and since the proportions of lengths are preserved by affine maps, then $f(b$-median line$)=B$-median line.

user126154
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  • you can also prove it directly, without use of affine by making use of similar triangles – user126154 Jul 14 '14 at 11:50
  • I want a mathematical proof I know we can use this but I wanna know the school proof – Apoorv Jain Jul 14 '14 at 13:18
  • what I gave is a mathematical proof, what do you mean by "I want a mathematical proof"? – user126154 Jul 14 '14 at 15:45
  • Actually I want a simple proof which is applicable at school level because the unique affine map I can't understand what it is. – Apoorv Jain Jul 14 '14 at 15:48
  • If you want a proof with "similar triangles" just drow any line L passing through $P$ and crossing both AC and XY in points s,t respectively. Then note that by using similar triangles you see in the picture, you can prove that the ratio As/sC is the same as Yt/tX (note the "switch" of the orders). Then note that the median line gives you As/sC=1. So it is an admissible line. This implies that $P$ passes trhoug the median line. Since this holds for any $P$ the locus of $P$'s is the median line – user126154 Jul 14 '14 at 15:52
  • Could you please be more generous and explain this answer it seems simple – Apoorv Jain Jul 14 '14 at 15:53
  • this is not a matter of generosity: is a lack of pictures! – user126154 Jul 14 '14 at 15:54
  • So you could just post a new answer please – Apoorv Jain Jul 14 '14 at 15:55
  • @user3650050 thanks for asking to write a new answer... – user126154 Jul 14 '14 at 17:33
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Let L be a line through $P$ that crosses both $AC$ and $XY$.

Let $S=L\cap AC$ and $T=L\cap XY$. Triangles $SCP$ and $PTY$ are similar, so are $ASP$ and $PXT$ and the ration of lengths is given by $SP/PT$.

In particular $$\frac{AS}{SC}=\frac{XT}{TY}$$

Since the ratio $AS/SC$ runs from $0$ to $\infty$ as $L$ change among the lines crossing $AC$ and $XY$, there is a line such that the ratio is $1$.

Now, choose $L$ so that such ratio is $1$. Then $L$ is the $B$-median of $ABC$. Therefore $P$ belongs to the $B$-median of $ABC$. But this holds true for any such $P$. Therefore the locus of $P$'s is contained in the $B$-median of $ABC$.

Conversely you easlily see that in fact the $P$'s cover the whole median line.

user126154
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