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#1 h Mar 3, 2011, 7:10 AM • 1 Y

Y by Adventure10

Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ prove that $\frac{CN}{IM} = 2$

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#2 h Mar 3, 2011, 1:23 PM • 2 Y

Y by Adventure10, Mango247

Using barycentric coordinates:
Parallels $l_a \parallel CA, l_b \parallel BC$ go through $A_1 = (0: b: c),$ $B_1= (c:0:a)$ and through infinite points $(-1:0:1), (0:1:-1)$ $\Longrightarrow$ their equations are $l_a:\ -cx + ay+az = 0,$ $l_b:\ bx-cy+bz = 0.$ They cut the line $CI$ with equation $bx-ay=0$ at $A_2=(a: b: c-a)$ and $B_2=(a:b:c-b).$ Equations of $AA_2, BB_2$ through $A=(1:0:0), B=(0:1:0)$ are then $AA_2:\ (c-a)y - bz = 0,$ $BB_2:\ (c-b)x -az = 0$ . The intersection $N \equiv AA_2 \cap BB_2$ is $N=(a(c-a):b(c-b):(c-a)(c-b))$ . If $CN \parallel IM,$ $N$ satisfies equation of the C-Nagel line $(c+a -b)x - (b+c-a)y = 0$ $\Longrightarrow$ $(a-b)(a^2 + b^2 - c^2) = 0$ $\Longrightarrow$ non-isosceles $\triangle ABC$ is right at $C.$ Expanding coordinates of $N$ by $a+b+c$ and using $a^2 + b^2 = c^2,$ they become $N = (b+c-a:c+a-b:a+b-c),$ hence $N$ is Nagel point of the right $\triangle ABC.$ Since incenter $I$ is Nagel point of the complementary (medial) triangle $\triangle KLM$ ( $K, L$ midpoints of $BC, CA$ ), it follows that $CN = 2\ IM.$

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