A Geometry Question Concerning Six Equilateral Triangles The question is as follows:

The figure below is built by joining six equilateral triangles ABC, ACD, CDE, DEF, EFG, and FGH, all of whose edges are 1 unit long. It is given that HIJKLMB is straight.
(a) There are five triangles in the figure that are similar to CMB. List them, making sure that you match corresponding vertices.
(b) Find the lengths of CM and EK.
(c) List the five triangles that are similar to AMB.
(d) Find the lengths of CL, HI, IJ, and JK.


I believe that we need to use a concept based on similar triangles, but I am unsure as to what I should be doing. Any help will be greatly appreciated.
 A: Since $\Delta CMB\sim\Delta AMH$, we obtain:
$$\frac{CM}{AM}=\frac{BC}{AH}$$ or
$$\frac{CM}{AC-CM}=\frac{BC}{AH},$$
which says $$CM=\frac{1}{4}.$$
$EK=\frac{1}{2}$ by symmetry.
Now, $\Delta BLC\sim\Delta HLD.$
Thus, $$\frac{CL}{LD}=\frac{BC}{HD}$$ or
$$\frac{CL}{1-CL}=\frac{1}{2},$$
which gives $$CL=\frac{1}{3}.$$
Now, since $FI=CM=\frac{1}{4}$, by the law of cosines we obtain:
$$HI=\sqrt{1^2+\left(\frac{1}{4}\right)^2-2\cdot1\cdot\frac{1}{4}\cdot\frac{1}{2}}=\frac{\sqrt{13}}{4}.$$
Also, since $FI=\frac{1}{2}$ and $AJ=\frac{1}{3}$, we obtain:
$$IJ=\sqrt{\left(\frac{1}{4}\right)^2+\left(\frac{1}{3}\right)^2-2\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}}=\frac{\sqrt{13}}{12}$$ and since $EJ=1-\frac{1}{3}=\frac{2}{3},$ we obtain:
$$JK=\sqrt{\left(\frac{2}{3}\right)^2+\left(\frac{1}{2}\right)^2-2\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}}=\frac{\sqrt{13}}{6}.$$ 
Done!
A: Here's the interesting part (to me):

$$|\overline{AP}|:|\overline{PQ}|:|\overline{QR}|:|\overline{AB}| = 3:1:2:12$$
A: As $BMC\sim BKE\sim BIG$ we have 
$$CM:KE:IJ=1:2:3.$$ 
But $KE=KD=1/2$ (by symmetry), then:
$$
CM=IF={1\over4},\quad IG=AM={3\over4}.
$$
In addition, $LMC\sim BMA$ and $CL:AB=CM:AM=1:3$, whence: $CL=1/3$.
Finally: $GHI\sim FJI\sim EJK$ and we know that $FJ=CL=1/3$ and $JE=2/3$, so we have:
$$
HI:IJ:JK=3:1:2,
\quad\hbox{which entails}\quad
IJ:HK=1:6
\quad\hbox{and}\quad
IJ:HB=1:12.
$$
so that:
$$
IJ={HB\over12},\quad
JK={HB\over6},\quad
IJ={HB\over4},
$$
and we can compute $HB=\sqrt{13}$ by Pythagoras' theorem to get the required lengths.
