It is known that the diagonal of a regular pentagon are in golden ratio to its sides. So we need to prove $BK:AK$ is golden ratio. Now $BK:AK=BL:AK=ML:MA$(Using similar triangles). Take $ML=b,MA=a$. Let $N$ be the midpoint of side $BC$. Then $NL=b-a$. By Power of Point -$BN.NC=NK.NL$ which simplifies to $b^2-a^2=ab$. Dividing by $a^2$ we get $ML:MA$ as golden ratio. And so $BK:AK$ is golden ratio. Stats - 14 students solved it in contest.

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Let $L$ be midpoint of $AB$. WLOG, let $AB=2$. Let $x=AK$ and $y=BK$. Note that $\triangle BKL \sim \triangle BAK$ so $BK^2=BL \cdot BA$; so $y=\sqrt{2}$. By cosine law in $\triangle AKB$, we get $$x^2+y^2+xy=4 \implies x=\frac{\sqrt{5}-1}{\sqrt{2}}.$$Thus, we conclude that $\tfrac{x}{y}$ is the golden ratio, as desired.

Here is a solution using the technique in the solution to the original version of TSTST 2018/5. Extend $\overline{BK}$ to meet $\overline{AC}$ at $X$, so $KB = KX$. Then $\triangle KBC \sim \triangle KAX$, so \[KA \cdot KC = KX \cdot KB = KB^2.\]Since $KA + KB = KC$, it follows that $\tfrac{KB}{KA} = \varphi$.

Solution using ptolmeys:- extend KM to meet the circumcircle at a point L not same as K. By symmetry $BL=BK=a$ and $AK=CL=b$ join $AL$ by ptolmeys theorom on quadrilateral $ABCL$ u get $AL=a+b$ . By similarity $\frac{b}{a}=\frac{KM}{MB}=\frac{KM}{MA}=\frac{a}{a+b}$ $\implies b^2+ab=a^2$ $\implies (\frac{b}{a})^2+\frac{b}{a}-1=0$ $\implies \frac{b}{a}=\frac{-1+\sqrt{5}}{2}$