We will use complex numbers, let WLOG the incircle be a unit circle, $A=a; B=b ;C=c; K=k; N=n; M=m$ Hence, $a=\frac{2km}{k+m} ; b=\frac{2kn}{k+n} ; c=\frac{2mn}{m+n}$ Since, $\angle ANM =\angle CKM$, Hence, $$\frac{\tfrac{a-n}{m-n}}{\tfrac{c-k}{m-k}} \in R \implies \frac{(a-n)(m-k)}{(m-n)(c-k)} = \overline{\left(\frac{(a-n)(m-k)}{(m-n)(c-k)}\right)} $$And since, $k,m,n$ lie on the unit circle, we can convert $a,b,c$ to $k,m,n$ and exploit their property of conjugates, $$ \implies \frac{(a-n)(m-k)}{(m-n)(c-k)}=\frac{n(m-k)}{k(m-n)} \left(\frac{\tfrac{\tfrac{2}{km}}{\tfrac{1}{k}+\tfrac{1}{m}}-\tfrac{1}{n}}{\tfrac{\tfrac{2}{mn}}{\tfrac{1}{m}+\tfrac{1}{n}}-\tfrac{1}{k}} \right) \implies \frac{a-n}{c-k}=\frac{(m+n)(2n-m-k)}{(m+k)(2k-m-n)} \implies \frac{\tfrac{2km}{k+m}-n}{\tfrac{2mn}{m+n}-k}= \frac{(m+n)(2n-m-k)}{(m+k)(2k-m-n)}$$ After simplifying this, $$4k^2m-2km^2-2kmn-2k^2n+kmn+kn^2-2kmn+m^2n+mn^2=4mn^2-2m^2n-2kmn-2kmn+km^2+k^2m-2kn^2+kmn+k^2n \implies \sum_{cyc} km(k-m) =0$$This is equivalent to, $(k-m)(m-n)(n-k)=0 \implies \text{ either } k=m ; k=n \text{ or } m=n$ which indicates after checking properly that atleast one of the three conditions satisfy and and all of these conditions lead to that $\Delta ABC$ is isosceles

parmenides51 wrote: The inscribed circle of the triangle $ABC$ touches its sides $AB, BC, CA$, at points $K,N, M$ respectively. It is known that $\angle ANM = \angle CKM$. Prove that the triangle $ABC$ is isosceles. (Vyacheslav Yasinsky) My solution: We know that $AN,BM,CK$ are concurrent at $I$. Let $AN,CK$ intersect the inscribed circle at the second points are $G,H$, respectively. It is easy to see that $GH \parallel AC$. $\Rightarrow \dfrac{CH}{AG}=\dfrac{IH}{GI}=\dfrac{NH}{GK} \Rightarrow \triangle NHC \sim \triangle KGA \Rightarrow \angle HNC=\angle GKA \Rightarrow NH=KG$. It implies $AK=NC$. Q.E.D

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