Let $ ABC$ be a triangle and $P$ a random point on the same plane as the triangle. Let $l$ be a line passing through $P$. Let $A_1,B_1,C_1$ be the intersection points of $BC,CA,AB$ with the reflections of $AP,BP,CP$ across line $l$ in respective order. Prove that $A_1,B_1,C_1$ are collinear.

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First reflect $\Delta ABC$ across $l$. Suppose the image is $A'B'C'$. Now notice that these two triangles are perspective from the point at infinity. Hence the corresponding sides intersect at points that are collinear. These collinear points lie on $l$. Furthermore, let $PA',PB',PC'$ intersect $BC,CA,AB$ at $A_1,B_1,C_1$ respectively. We have to show that these three points are collinear. After that, I have tried to look for more perspective triangles but haven't been able to find anything that helps me.

How do I proceed? Am I overthinking?

Thanks in advance.

  • $\begingroup$ Can you put an image I don't coudn't draw it :P thanks $\endgroup$ – Weijie Chen Mar 30 '16 at 22:06
  • $\begingroup$ @WeijieChen Done. $\endgroup$ – rah4927 Mar 31 '16 at 8:21

By applying a projective transform that moves $P$ to infinity and preserves symmetry about the line $\ell$, it is sufficient to prove the statement in the limit case when $P$ is the ideal point of $\ell$.

In the limit case the lines $A'A_1$, $B'B_1$ and $C'C_1$ are parallel to $\ell$. For every point $X$, let $d(X)$ denote the signed distance between $\ell$ and $X$. Let $x=d(A)$, $y=d(B)$ and $z=d(C)$; then $d(A')=d(A_1)=-x$, $d(B')=d(B_1)=-y$ and $d(C')=d(C_1)=-z$. Hence, $$ \frac{AC_1}{C_1B} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A} = \frac{x+z}{-z-y} \cdot \frac{y+x}{-x-z} \cdot \frac{z+y}{-z-x} = -1, $$ so $A_1,B_1,C_1$ are collinear by Menelaus.

It is possible to do the same without moving $P$ into infinity; in that case you have to work with cross-ratios of the lines $\ell$, $PA$, $PA'$, etc.


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