I think $\angle C =90$ ... And $l_{a}, BN, IM$ are concur at a point for any ABC.

Little Gauss wrote: I think $\angle C =90$ ... And $l_{a}, BN, IM$ are concur at a point for any ABC. Please show some proofs

Anybody?

Little Gauss wrote: $l_{a}, BN, IM$ are concur at a point for any ABC. Why is this? Is it well-known?

Hmm... please someone show how to prove that, because I don't think it's well-known.

My solution does not need that property. I have a ugly proof. First I show that $ CN\parallel IM \Rightarrow a^{2}+b^{2}=c^{2}$. Next, I calculate the ratio $ \frac{CN}{IM}$ in case $ \angle C=90$. The answer is $ 2$.

Little Gauss wrote: First I show that $ CN\parallel IM \Rightarrow a^{2}+b^{2}=c^{2}$. Next, I calculate the ratio $ \frac{CN}{IM}$ in case $ \angle C=90$. I can't see how to do it. In fact, no matter how I draw the figure, even if I make $ \angle C = 90^{\circ}$, $ CN$ and $ IM$ are still not parallel!!

Can you (or anyone) post the detailed complete solution?

I do a very large but interesting solution i think, here is: Lemma: If X is the intersection of A_1A_2 and IM, then B_2, X and B are collinear. Proof: Let N the midpoint of BC, R is the intersection oof CI and AB, T in AC such that RT is parallel to BC, IL and IS are parallel to AC and BC respectively and those point are on BC and AC. $ \frac{LA_1}{SB_1}=\frac{CA_1.a}{CB_1.b}$ We have that: CR, BB_1 and AA_1 concur, AN, BT and CR too. Y is the intersection of TN and AB, Y_1 is the intersection of A_1B_1 and AB, then $ (A,B,R,Y)=-1=(A,B,R,Y_1)$, then $ Y=Y_1$ By menelaus theorem for 1)triangle CTN and line B_1A_1 2)triangle CTN and line AB, get: $ \frac{A_1N}{B_1T}=\frac{A_1C}{B_1C}.\frac{NB}{AT}.\frac{b}{a}$ then: $ \frac{IX}{XM}.\frac{MB}{BR}.\frac{RB_2}{B_2I}=\frac{LA_1}{B_1S}.\frac{MB}{BR}.\frac{TB_1}{A_1N}$ $ =\frac{a^2}{b^2}.\frac{MB}{BR}.\frac{AT}{BN}=\frac{a^2.MB}{b^2.BR}.\frac{AR.AC/AB}{MB.CB/AB}$ $ =\frac{a.AR}{b.BR}=1$, then for Menelaus theorem again, the lemma is proved. Returning to the problem: X and Y are the intersections of A_1A_2, BB_2 and B_1B_2, AA_2, for the lemma I,X,Y,M ARE COLLINEAR. Let S a point on CI, such that BS parallel to BN, note the triangles $ B_2A_2Y$ and $ CSB$ are similar, also triangles: B_2XY and BNC, then $ SB/A_2Y=CB/B_2Y$, $ AN/CN=A_2Y/XY$ and $ B_2Y/XY=CB/CN$, from this: $ AN=SB$ then SANB is a parallelogram with center M, finally S,M,N are collinear. M is mdpoint of SN, and IM is parallel to CN: $ \frac{CN}{IM}=2$ please tell me if you dont understand a step. Little Gauss, can you please post your solution?

I've made a figure. Can it help ??

Attachments:

cuenca wrote: $ \frac {LA_1}{SB_1} = \frac {CA_1.a}{CB_1.b}$ I don't understand from the first step. Why is this?

To $ nguyentrang$ we have that the lines $ IL$ and $ A_1A_2$ are parallel so $ \frac {A_1L}{CA_1} = \frac {A_2I}{IC}$ and because the lines $ A_1A_2$ and $ AC$ are parallel: $ \frac {A_2I}{CI} = \frac {IA_1}{IA} = \frac {a}{a + b + c}$ then $ \frac {A_1L}{CA_1} = \frac {a}{a + b + c}$ similarly: $ \frac {SB_1}{CB_1} = \frac {b}{a + b + c}$, the result follows.

Little Gauss wrote: $ CN\parallel IM\ \Longleftrightarrow\ a^{2} + b^{2} = c^{2}$ . Next, I calculate the ratio $ \frac {CN}{IM}$ in case $ CA\perp CB$ . The answer is $ 2$ . Yes, and my proof (I"ll present it later) follows the same way.

bilarev wrote: Let $ w = C(I)I$ be the incircle of non-isosceles triangle $ ABC$ . Denote $ A_{1}\in AI\cap BC$ and $ B_{1}\in BI\cap AC$ . Let $ l_{a}$ be the line for which $ A_{1}\in l_a\parallel AC$ and $ l_{b}$ be the line for which $ B_{1}\in l_b\parallel BC$ . Denote $ A_2\in l_{a}\cap CI$ , $ B_2\in l_{b}\cap CI$ , $ N\in AA_{2}\cap BB_{2}$ and the midpoint $ M$ of $ AB$ . If $ CN\parallel IM$ find $ \frac {CN}{IM}$ . Proof. Denote $ \|\begin{array}{c} X\in AN\cap BC \\ \ Y\in BN\cap CA \\ \ Z\in CN\cap AB\end{array}$ . Observe that $ A_1C = \frac {ab}{b + c}$ , $ B_1C = \frac {ab}{a + c}$ , $ C_1M = \frac {c(a - b)}{2(a + b)}$ $ \implies$ $ \blacktriangleright\ A_1A_2\parallel AC\Longrightarrow A_1A_2 = A_1C\Longrightarrow$ $ \frac {XA_1}{XC} = \frac {A_1A_2}{AC} = \frac {A_1C}{AC} = \frac {IA_1}{IA} = \frac {a}{b + c}$ $ \implies$ $ \frac {XA_1}{a} = \frac {XC}{b + c} =$ $ \frac {A_1C}{b + c - a} = \frac {ab}{(b + c)(b + c - a)}$ $ \implies$ $ XC = \frac {ab}{b + c - a}\ ,\ XB = \frac {a(c - a)}{b + c - a}$ $ \implies$ $ \boxed {\ \frac {XB}{XC} = \frac {c - a}{b}\ }$ . $ \blacktriangleright\ B_1B_2\parallel BC\Longrightarrow B_1B_2 = B_1C\Longrightarrow$ $ \frac {YB_1}{YC} = \frac {B_1B_2}{BC} = \frac {B_1C}{BC} = \frac {IB_1}{IB} = \frac {b}{a + c}$ $ \implies$ $ \frac {YB_1}{b} = \frac {YC}{a + c} =$ $ \frac {B_1C}{a + c - b} = \frac {ab}{(a + c)(a + c - b)}$ $ \implies$ $ YC = \frac {ab}{a + c - b}\ ,\ YA = \frac {b(c - b)}{a + c - b}$ $ \implies$ $ \boxed {\ \frac {YA}{YC} = \frac {c - b}{a}\ }$ . Apply the Menelaus' theorem to the transversals : $ \blacktriangleright$ $ \overline {NBY}$ in $ \triangle\ AXC\ :$ $ \frac {BX}{BC}\cdot\frac {YC}{YA}\cdot\frac {NA}{NX} = 1$ $ \implies$ $ \frac {NX}{NA} = \frac {c - a}{b + c - a}\cdot \frac {a}{c - b}$ . $ \blacktriangleright$ $ \overline {CNZ}$ in $ \triangle\ ABX\ :$ $ \frac {CX}{CB}\cdot\frac {ZB}{ZA}\cdot\frac {NA}{NX} = 1$ $ \implies$ $ \frac {ZA}{b(c - b)} = \frac {ZB}{a(c - a)} = \frac {c}{a(c - a) + b(c - b)}$ . Observe that $ MZ = AZ - AM = \frac {bc(c - b)}{a(c - a) + b(c - b)} - \frac c2$ $ \implies$ $ MZ = \frac {c(a - b)(a + b - c)}{2[a(c - a) + b(c - b)]}$ . Therefore, $ \boxed {\ IM\parallel CN\ }$ $ \Longleftrightarrow$ $ IM\parallel CZ$ $ \Longleftrightarrow$ $ \frac {MC_1}{MZ} = \frac {IC_1}{IC}$ $ \Longleftrightarrow$ $ \frac {c(a - b)}{2(a + b)}\cdot\frac {2[a(c - a) + b(c - b)]}{c(a - b)(a + b - c)} =$ $ \frac {c}{a + b}$ $ \Longleftrightarrow$ $ c(a + b - c) = a(c - a) + b(c - b)$ $ \Longleftrightarrow$ $ \boxed {\ a^2 + b^2 = c^2\ }$ . Remark. Let $ [UV]$ be the diameter of the incircle, where $ U\in BC$ . From the well-known property " $ CV$ is tha $ C$ - Nagel's line and $ MI\parallel CV$ " obtain that $ \overline {CNZ}$ is the Nagel's line, i.e. $ BZ=p-a$ . In this case $ BZ = \frac {ac(c - a)}{a(c - a) + b(c - b)} = \frac {ac(c - a)}{ac + bc - c^2}$ $ \implies$ $ BZ = \frac {a(c - a)}{a + b - c}$ . Observe that $ BZ = p - a$ $ \Longleftrightarrow$ $ \frac {a(c - a)}{a + b - c} = p - a$ $ \Longleftrightarrow$ $ 2a(c - a) = (a + b - c)(b + c - a)$ $ \Longleftrightarrow$ $ 2a(c - a) = b^2 - (a - c)^2$ $ \Longleftrightarrow$ $ 2a(c - a) = b^2 - a^2 - a^2 - b^2 + 2ac$ what is truly. Therefore, the line $ CZ$ is the $ C$ - Nagel's line in $ \triangle\ ABC$ . Prove easily analogously that $ XB = YA = p - c$, i.e. the point $ N$ is the Nagel"s point of $ \triangle\ ABC$ . Apply the Menelaus' theorem to the transversal $ \overline {BNB_2}$ in $ \triangle\ ZCC_1\ : \ \frac {BZ}{BC_1}\cdot\frac {B_2C_1}{B_2C}\cdot\frac {NC}{NZ} = 1$ $ \implies$ $ \frac {a(c - a)}{a + b - c}\cdot\frac {a + b}{ac}\cdot\frac {c - b}{a + b}\cdot \frac {NC}{NZ} = 1$ $ \implies$ $ \frac {NZ}{(c - a)(c - b)} = \frac {NC}{c(a + b - c)} = \frac {CZ}{ab}$ . In conclusion, $ \frac {CZ}{IM} = \frac {CC_1}{IC_1}$ and $ \frac {CN}{IM} = \frac {CN}{CZ}\cdot\frac {CZ}{IM} =$ $ \frac {c(a + b - c)}{ab}\cdot \frac {a + b + c}{c} =$ $ \frac {(a + b - c)(a + b + c)}{ab} = \frac {(a + b)^2 - c^2}{ab} = 2$ $ \implies$ $ \boxed {\ \frac {CN}{IM} = 2\ }$ .