Problem
1982 AHSME Problem 24
In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If AG=2, GF=13, FC=1, and HJ=7, then DE equals
![[asy] defaultpen(fontsize(10)); real r=sqrt(22); pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J); dot(A^^B^^C^^D^^E^^F^^G^^H^^J); draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7)); label("$A$", A, N); label("$B$", B, dir(210)); label("$C$", C, dir(330)); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, dir(170)); label("$G$", G, dir(250)); label("$H$", H, SE); label("$J$", J, dir(0)); label("2", A--G, dir(30)); label("13", F--G, dir(180+30)); label("1", F--C, dir(30)); label("7", H--J, dir(-30));[/asy]](/static/aops/1b3a1c5f657f7bea77cc84231dbb2e245d59d39c.png)
\text {(A)} 2\sqrt{22} \qquad \text {(B)} 7\sqrt{3} \qquad \text {(C)} 9 \qquad \text {(D)} 10 \qquad \text {(E)} 13
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