in the triangle ABC on the AC side, points M and N are chosen such that ABM = MBN = NBC in the triangle ABC on the AC side, points M and N are chosen such that &lt;ABM = &lt;MBN = &lt;NBC It turned out that NB = BC. On the side AB, a point K was marked such that BK = BM.  Prove that AK + NC&gt; AM.
I tried to get a triangle with sides AK NC and AM.
But I couldn't. So I don't know how to prove this
Geometric hint:
Consider the figure:
Following cases can be considered:
a-If BC=BF then triangle BCF is isosceles then NC=DE and  FD+DE&gt;FE. This is a particular case.
b-If BC&lt;AB then we have:
AK+KG=AK+DE&gt;AM
Since $\angle AKM&gt;90^o$ and:
$NC&gt;DE&gt;\frac{KM}{2}$
Therefore:
AK+NC&gt;AM
let $BC=a$ and $\measuredangle ABC=3\beta$.
Thus, $$\measuredangle ACB=90^{\circ}-\frac{\beta}{2},$$
$$\measuredangle ANC=90^{\circ}+\frac{\beta}{2},$$
$$\measuredangle AMB=90^{\circ}+\frac{3\beta}{2}$$ and
$$\measuredangle BAC=180^{\circ}-\left(90^{\circ}+\frac{5\beta}{2}\right),$$ which says
$$90^{\circ}+\frac{5\beta}{2}&lt;180^{\circ},$$ which gives $$0^{\circ}&lt;\beta&lt;36^{\circ}.$$
Now,  $$NC=2a\sin\frac{\beta}{2}$$ and by law of sines we can show that:
$$AM=\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ and
$$AK=AB-BK=\frac{a\cos\frac{\beta}{2}}{\cos\frac{5\beta}{2}}-\frac{a\cos\frac{\beta}{2}}{\cos\frac{3\beta}{2}}=\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}.$$
Id est, we need to prove that:
$$\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+2a\sin\frac{\beta}{2}&gt;\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or
$$\frac{\cos\frac{\beta}{2}\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+1&gt;\frac{\cos^2\frac{\beta}{2}}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or $$2\cos\frac{\beta}{2}\sin2\beta+\cos4\beta+\cos\beta&gt;1+\cos\beta$$ or
$$\cos\frac{\beta}{2}\sin2\beta&gt;2\sin^22\beta$$ or
$$1&gt;4\sin\frac{\beta}{2}\cos\beta$$ or
$$1&gt;4\sin\frac{\beta}{2}\left(1-2\sin^2\frac{\beta}{2}\right)$$ or
$$1-2\sin\frac{\beta}{2}&gt;2\sin\frac{\beta}{2}\left(1-4\sin^2\frac{\beta}{2}\right)$$
$$1&gt;2\sin\frac{\beta}{2}\left(1+2\sin\frac{\beta}{2}\right)$$ or
$$\sin\frac{\beta}{2}&lt;\frac{\sqrt5-1}{4},$$ which we got by looking for  domain of $\beta$.