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 <ABM = <MBN = <NBC It turned out that NB = BC. On the side AB, a point K was marked such that BK = BM.  Prove that AK + NC> 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
 A: 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>FE. This is a particular case.
b-If BC<AB then we have:
AK+KG=AK+DE>AM
Since $\angle AKM>90^o$ and:
$NC>DE>\frac{KM}{2}$
Therefore:
AK+NC>AM
A: 
Let $|CH|=|HN|$, so $BH\perp AC$
and $|BH|=h$ is the height of
$\triangle ABC$,
$\triangle MBC$,
$\triangle NBC$.
Also let $|BC|=a=|BN|$,
$|BM|=d=|BK|$.
\begin{align} 
a&=\frac h{\cos\phi}
,\\
|NC|&=2h\tan\phi
.\\
d&=\frac h{\cos 3\phi}
,\\
|AK|&=
\frac h{\cos 5\phi}-d
=
h\cdot\frac{\cos 3\phi-\cos 5\phi}{\cos 3\phi \cos 5\phi}
,\\
|AM|&=
h\,(\tan 5\phi-\tan 3\phi)
\\
&=
\frac{8h\sin^2\phi\cos 2\phi}{\cos\phi(2\sin\phi+1)}
\cdot
\frac 1{1+2\sin\phi-4\sin^2\phi}
,\\
\end{align}
Since $\sin\phi,\cos\phi,\cos2\phi$ are positive
for $\phi\in(0,\tfrac\pi6)$
the question simplifies to
\begin{align} 
1+2\sin\phi-4\sin^2\phi
&>0
\quad \text{for } \phi\in(0.\tfrac\pi3)
.
\end{align}
A: 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
$$90^{\circ}+\frac{5\beta}{2}<180^{\circ},$$ which gives $$0^{\circ}<\beta<36^{\circ}.$$
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:
$$\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+2a\sin\frac{\beta}{2}>\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or
$$\frac{\cos\frac{\beta}{2}\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+1>\frac{\cos^2\frac{\beta}{2}}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or $$2\cos\frac{\beta}{2}\sin2\beta+\cos4\beta+\cos\beta>1+\cos\beta$$ or
$$\cos\frac{\beta}{2}\sin2\beta>2\sin^22\beta$$ or
$$1>4\sin\frac{\beta}{2}\cos\beta$$ or
$$1>4\sin\frac{\beta}{2}\left(1-2\sin^2\frac{\beta}{2}\right)$$ or
$$1-2\sin\frac{\beta}{2}>2\sin\frac{\beta}{2}\left(1-4\sin^2\frac{\beta}{2}\right)$$
$$1>2\sin\frac{\beta}{2}\left(1+2\sin\frac{\beta}{2}\right)$$ or
$$\sin\frac{\beta}{2}<\frac{\sqrt5-1}{4},$$ which we got by looking for  domain of $\beta$.
