How is the lettering possible? Does anyone have a diagram?

[asy][asy] import graph; size(8cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.300000000000004,xmax=11.16000000000001,ymin=-2.960000000000005,ymax=6.300000000000007; draw((1.340000000000002,3.520000000000004)--(0.08000000000000008,3.140000000000004)); draw((0.08000000000000008,3.140000000000004)--(-0.6409567066490278,2.038990723409752)); draw((-0.6409567066490278,2.038990723409752)--(-0.4855264643647211,0.7321466471009354)); draw((-0.4855264643647211,0.7321466471009354)--(0.4735631890733660,-0.1690453616674585)); draw((0.4735631890733660,-0.1690453616674585)--(1.787543545727366,-0.2429075463591564)); draw((1.787543545727366,-0.2429075463591564)--(2.841588593453857,0.5451210301298309)); draw((2.841588593453857,0.5451210301298309)--(3.142498940015772,1.826313038898225)); draw((3.142498940015772,1.826313038898225)--(2.549475290010800,3.001184500180175)); draw((2.549475290010800,3.001184500180175)--(1.340000000000002,3.520000000000004)); draw((1.340000000000002,3.520000000000004)--(2.299089653438090,2.618807991231610)); draw((2.299089653438090,2.618807991231610)--(1.998179306876175,1.337615982463216)); draw((1.998179306876175,1.337615982463216)--(0.7381793068761738,0.9576159824632160)); draw((0.7381793068761738,0.9576159824632160)--(-0.2209103465619136,1.858807991231610)); draw((-0.2209103465619136,1.858807991231610)--(0.08000000000000008,3.140000000000004)); draw((3.142498940015772,1.826313038898225)--(2.299089653438090,2.618807991231610)); draw((2.841588593453857,0.5451210301298309)--(2.299089653438090,2.618807991231610)); draw((0.7381793068761738,0.9576159824632160)--(0.4735631890733660,-0.1690453616674585)); draw((1.998179306876175,1.337615982463216)--(0.4735631890733660,-0.1690453616674585)); dot((1.340000000000002,3.520000000000004),ds); label("$A$",(1.420000000000002,3.640000000000004),NE*lsf); dot((0.08000000000000008,3.140000000000004),ds); label("$B$",(0.1600000000000002,3.260000000000003),NE*lsf); dot((-0.6409567066490278,2.038990723409752),ds); label("$C$",(-0.5600000000000007,2.160000000000002),NE*lsf); dot((-0.4855264643647211,0.7321466471009354),ds); label("$D$",(-0.4000000000000005,0.8600000000000000),NE*lsf); dot((0.4735631890733660,-0.1690453616674585),ds); label("$E$",(0.5600000000000007,-0.04000000000000119),NE*lsf); dot((1.787543545727366,-0.2429075463591564),ds); label("$F$",(1.860000000000002,-0.1200000000000013),NE*lsf); dot((2.841588593453857,0.5451210301298309),ds); label("$G$",(2.920000000000003,0.6599999999999998),NE*lsf); dot((3.142498940015772,1.826313038898225),ds); label("$H$",(3.220000000000004,1.940000000000002),NE*lsf); dot((2.549475290010800,3.001184500180175),ds); label("$I$",(2.620000000000003,3.120000000000003),NE*lsf); dot((-0.2209103465619136,1.858807991231610),ds); label("$J$",(-0.1200000000000001,1.880000000000001),NE*lsf); dot((0.7381793068761738,0.9576159824632160),ds); label("$K$",(0.8200000000000010,1.080000000000000),NE*lsf); dot((1.998179306876175,1.337615982463216),ds); label("$L$",(1.760000000000002,1.460000000000001),NE*lsf); dot((2.299089653438090,2.618807991231610),ds); label("$M$",(1.920000000000002,2.480000000000002),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

@above Thanks, but the lettering of the hexagon is counterclockwise instead of clockwise as stated in the problem.

Rukevwe wrote: @above Thanks, but the lettering of the hexagon is counterclockwise instead of clockwise as stated in the problem. I saw that. The problem barely makes sense in that way.

I think this problem has got to do with similar triangles and parallelograms.

$\tan(\angle HMG)=\frac{\sqrt{3}\cos 10^{\circ}}{\sqrt{3}+\cos 10^{\circ}}=\tan(\angle KEL)$

First, notice that the angles of $ABCDEFGHI$ are all equal to $140$ and the angles of $ABJKLM$ are $120$. Also, $AB= BC= CD= DE= EF= FG= GH= HI= IA= AM= ML= LK= KJ (*)$ Then, see that $A, M, H$ look collinear? Indeed, they are, and the prove is very simple. This happens because $\angle MAI= \angle BAI- \angle MAB= 140-120= 20$. So, as $AM= AI\implies \angle AMI= \angle AIM= 80$. $\Rightarrow \angle MIH= \angle AIH- \angle AIM= 140-80= 60$. Then, we know that $\angle IHA= \angle IAH= 20 \implies \angle IAH= \angle IAM$ $\implies A, M, H$ collinear. Similarly, we obtain $B, J, D$ collinear. Now, let's see another points collinear! We know that $\angle HGB= 60$, and as $\angle AHG= \angle IHG-\angle IHM= 140-20= 120\implies GB//AM$, but $BL//AM$ $\implies B,L, G$ are collinear and still stands the parallelism: $BL//MH$ (1) same for $A, K, E$ collinear and the parallelism: $EK//DJ$. (2) By (*), we know that $LM= GH= JK= DE$. (3) Then, by (1) and (2) with (3): $\rightarrow MHGL$ parallelogram $\implies MH= GL$(4) $\rightarrow JKED$ parallelogram $\implies JD= KE$ (5) Now, we could have 2 endings: First ending: $AIH\cong BCD (SAS)\implies AH= BD\implies AM+MH= BJ+JD\implies MH= JD$ (by (*), $AM= BJ$) By (4) and (5), $MH= GL= JD= KE\implies MH= KE\implies GHM\cong LKE (SAS)\therefore \angle HMG= \angle KEL$. Second ending: As $BL= AK$ and $BG= AE\implies KE= LG= MH\implies GHM\cong LKE (SAS)\therefore \angle HMG= \angle KEL$, as wanted.

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