Which state?

TheHimMan wrote: Which state? Idk It’s of KV

Ok,do you have the other problems as well?

You need not latex them r.n.

Let $CK = x$ and $CN = y.$ Then $CL = x+y.$ Now, applying Law of Cosines in $\Delta CNK$ gives \begin{align*} NK &= \sqrt{CK^2+CN^2-2\cdot CK\cdot CN\cdot \cos{\angle NCK}} \\ &= \sqrt{x^2+y^2-2xy\cos{120^{\circ}}} \\ &= \sqrt{x^2+y^2+xy} \end{align*}Similarly, by applying Law of Cosines in $\Delta CLK$ and $\Delta NCL$ we can show that $$ KL=LN=\sqrt{x^2+y^2+xy}.$$Hence, $\Delta KLN$ is equilateral. $QED.$

Let LS parallel AC and intersects BC at S.It’s not difficult to see the congruence of triangle LNC and triangle LKS.Triangle CLS is an equilateral so triangle KLN is also an equilateral.

very easy

Let $P \in CL \ni CK = CP.$ Then $$PL=CL-CP=CL-CK=CN.$$And the triangle $CPK$ is equilateral. So, $PK=CK$ and $\angle LPK = 120^{\circ}.$ Thus, by ASA congruence, $$\Delta LPK \cong \Delta NCK.$$So, $NK=LK$ and $\angle NKL = \angle NKP + \angle PKL = \angle NKP + \angle CKN = \angle CKP = 60^{\circ}.$ Hence, the triangle $KLN$ is equilateral. $QED.$

Need to prove $NLKC$ is a cyclic quad. Introduce point $X$ on $CL$ such that $CX = CK$. Then by ASA congruency, $$\Delta CNK \cong \Delta XLK.$$Hence, $ \angle CLK = \angle CNK = 60^{\circ} $. So $NLKC$ is cyclic.

Synthetic solution: In the figure, let $K'$ be a point on side $BC$ such that $CNLK'$ is cyclic. Therefore, $\angle K'CL=\angle K'NL=60^{\circ}$. $\angle NCL=\angle NK'L=60^{\circ}\implies K'NL$ is equilateral. Now, we prove that $K=K'$. Firstly, let $N'$ be a point on line $BC$ such that $CN=CN'$. Also, $\angle NCN'=60^{\circ}$. Therefore, $NCN'$ is equilateral. $\implies \angle CN'N=60^{\circ}=\angle NCL$. Also, as $CNLK'$ is cyclic, $\angle NLC=\angle NK'C= \angle NK'N'$. Therefore, by $AAS$ congruency criterion, $\Delta NN'K' \cong \Delta NCL \implies CN=N'K' = N'C+CK'= NC+CK'$. Therefore, $K=K'$, and we're done!

Attachments:

Let $CK=a$, and $CN=b$. Using cosine law, we see that $LK=LN=NK=\sqrt{a^2+ab+b^2}$