BdMO National Higher Secondary 2011/5

[BdMO]

Problem 5:
In a scalene triangle $ABC$ with $\angle A = 90^{\circ}$, the tangent line at $A$ to its circumcircle meets line $BC$ at $M$ and the incircle touches $AC$ at $S$ and $AB$ at $R$. The lines $RS$ and $BC$ intersect at $N$ while the lines $AM$ and $SR$ intersect at $U$. Prove that the triangle $UMN$ is isosceles.

[FahimFerdous]

Ok, I'm not so good at writing solutions in english, especially in geometry, so forgive me if there are any mistake. And I'll write some things in Bangla as I don't know the English words for them. Here's the solution:

As I'm writing in mobile, I will use '<' sign instead of 'angle' sign.
Suppose, $XM$ is the tangent at point $A$.
Now, $\angle AUR = \angle MUN$......(1) (বিপ্রতীপ কোণ) and $ \angle ASR = \angle CSN$......(2) (same as above).
$\angle ACB=\angle BAX$ (একান্তর বৃত্তাংশস্থ কোণ).
$\angle BAX + \angle BAM$ =180 degree=$\angle ACB+\angle ACM$.
So, $\angle BAM=\angle ACM$
or, $\angle RAU=\angle SCN$.......(3)
Again, $AR=AS$ (both are tangents to the incircle of triangle ABC).
So, $\angle ARS=\angle ASR$
or, $\angle ARU=\angle CSN$.......(4) (using (2))
Now, using (3) & (4) between $\triangle AUR$ and $\triangle CSN$ we can say, $\angle AUR = \angle SNC$
or, $\angle MUN=\angle MNU$ (using (1))
So, $MN=MU$
Hence, the $\triangle UMN$ is isolesces.

Is my solution correct? :-s

[Mehfuj Zahir]

I think it more easily.<ARS=<UNM+<ABC
<ASR=<SAU+<MUN(<MUN=<SUA)
<ASR=<ARS
so<UNM=<MUN [<ABC=<SAU by alternate segment theorem ]
MU=MN so,MUN triangle is isocles.

[FahimFerdous]

Hmm, nice.

[Labib]

Hmmm.....
Both of these solutions make the problem look so easy....
Hats off to you two...!!
I had a rather complicated solution. Now, seeing these two solutions and moderating my one, it seems much easier. As it resembles your solutions, I decide not to give it anymore.