We draw the Incircle of triangle $ABC$ that is tangent to $AB,BC$ at points $N,M$.We pass the line $l$ from the point $I$(Center of the incircle).Then we draw a perpendicular from $C,A$ to that line that intersect that line at $Y,X$.If $MY$ intersects the incircle at $D$ Prove that we can draw a line that contains $N,X,D$.

It seems that we should connect both $N,D$ to $X$ and the continue with angles because we have two right angles and can have more using tangents and we are able to use cyclic quadrilaterals.But I can't get anything using that.

  • $\begingroup$ Is $l$ any line passing through $I$, or is some description missing? $\endgroup$ – Matthew Conroy Aug 24 '17 at 6:55
  • $\begingroup$ @MatthewConroy It is a random line. $\endgroup$ – Taha Akbari Aug 27 '17 at 12:22

Besides M and N, let K be the third contact point of the in-circle to the line AC. Then, BMIN, ANIK and CMIK are cyclic.

enter image description here

In addition, AXIK is also cyclic. Thus, A, N, X, I, and K are points on the same (green) circle. Then, $\alpha = \alpha_2 =\alpha_1 =\alpha’$.

Similarly, C, M, I, K, and Y are con-cyclic points on the (blue) circle. Then, $\beta’ = \beta’’$.

Note that, by “angle in alternate segment”, $\beta = \beta’’$. Therefore, $\beta =\beta' = \beta''$.

Note also that $\angle 1 = \angle 2 = \theta + \beta = \theta + \beta’ = \angle DKY$. This means the line $\lambda$ is the perpendicular bisector of KD. Therefore, $\omega = \alpha’ = \alpha$.

Required result follows.

  • $\begingroup$ Can you upload the picture another where?My computer can't open pictures on SE sites. $\endgroup$ – Taha Akbari Aug 29 '17 at 8:18
  • $\begingroup$ @TahaAkbari Try postimg.org/image/45u1xf51h. $\endgroup$ – Mick Aug 29 '17 at 8:24
  • $\begingroup$ again doesn't work can you email it? $\endgroup$ – Taha Akbari Aug 29 '17 at 8:25
  • $\begingroup$ taakda22@gmail.com $\endgroup$ – Taha Akbari Aug 29 '17 at 8:26
  • $\begingroup$ @TahaAkbari image sent. $\endgroup$ – Mick Aug 29 '17 at 8:32

enter image description here

$$\angle NAI=\angle NXI=\alpha\ , \ \angle MCI=\angle MYI=\beta$$

$$\angle NDM=\angle NKN=\alpha+\beta \Rightarrow N,X,D \rightarrow collinear$$

  • $\begingroup$ 1) I guess the con-cyclicies of (ANIXK) and (CMIKY) are assumed. 2) $\angle NKI = \alpha$ too and similarly $\angle MKI = \beta$. 3) $\angle NKN$ should be $\angle NKM$ instead. $\endgroup$ – Mick Aug 24 '17 at 15:04
  • $\begingroup$ where is $K$??? $\endgroup$ – Taha Akbari Aug 25 '17 at 8:23
  • $\begingroup$ Can you upload the picture another where?My computer can't open pictures on SE sites. $\endgroup$ – Taha Akbari Aug 25 '17 at 11:29
  • $\begingroup$ @TahaAkbari savepic.net/9818539.png $\endgroup$ – Alex Ravsky Aug 25 '17 at 15:48
  • $\begingroup$ OMG.What was the motivation? $\endgroup$ – Taha Akbari Aug 28 '17 at 6:44

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