.....Emm,my geogebra shows this conclusion is wrong...

Does regular triangle mean an equilateral triangle?

PDT2020 wrote: Does regular triangle mean an equilateral triangle? Yes it does.

ironball wrote: .....Emm,my geogebra shows this conclusion is wrong... Because the question must be wrong indeed.

Mindless cart bash(very abridged): Set $A = (0,0),\hspace{0.01cm}K = (k,0),\hspace{0.01cm}N = (n,0)$ and $B = (a,0)$. Clearly, $C = \left (\frac{a}{2},\frac{\sqrt{3}a}{2} \right )$. Since $CL=AK$ and $CM=BN$ we get, after some deliberation, that $M = \left (a- \frac{n}{2},\frac{\sqrt{3}n}{2} \right )$ and $L = \left (\frac{a-k}{2},\frac{\sqrt{3}(a-k)}{2} \right )$. The condition $ML=KN$, again after some efforts, yields $3nk+a^2 = 2an+ak$. $KL \parallel MN$ means that the slope of $\overleftrightarrow{KL}$ is the same as $\overleftrightarrow{MN}$ which can be checked to reduce down to the above equation. Proved.

This is the solution I've made in the exam.

Attachments:

Outline. Let $P$ be point on $BC$ so that $BP=CL=AK$ and let $Q$ be point on $AC$ so that $AQ=CM=BN$. Now notice that $\triangle AQK\cong\triangle CML\cong\triangle BNP$, since they all have sixty degree angle and two equal side lengths. Hence, we have $PN=KQ=LM=KN=QL=PM$. Extend $NM$ to hit $AC$ at $R$. We have $$\angle ALK=\frac{\angle AQK}{2}=\frac{\angle BNP}{2}=\frac{120^\circ-\angle BPN}{2}=60^\circ-\angle BMN=\angle ARN\quad\square$$

My solution is essentially the same as above. We just mark points on the other two sides which help us to translate the side equality condition into something more manageable.

Seems that my solution is different so I will just post it here. We know, $$AB=AC$$$$AK+KB=AL+LC$$$$KB=AL$$Simlarly we get $BM=AN$ we see that in $\triangle{ALN}$ and $\triangle{BKM}$ \begin{align} LA=KB \\ \angle{LAN}=\angle{KBM} & \text{ since all angles in a regular$\triangle{}$ are 60 }\\ AN=BM \end{align}therefore the $\triangle ALN\cong\triangle BKM$ So $KM=NL$ By SSS in $\triangle{KNM}$ and $\triangle{LMN}$ We get $$\angle{NKM}=\angle{NLM}$$so quadrilateral $KNML$ is cyclic. and has the sides $NK=ML$. which implies it is an isoceles trapezoid; since a cyclic quadrilateral with two opposite sides equal have the other two sides parallel. implying sides $KL$ and $ML$ are parallel.