A Ratio Of Shapes

Draw a line GF, that GF // ED.

∠GFE = ∠CEF (alt. ∠s, GF // ED)

∠GBF = ∠CBE (vert. opp. ∠s)

FB = BE (given)

∴ △BFG≅△BEC (ASA)

Let GB be k.

BC = GB = k (proved)

AG = GC = 2k (mid-point theorem)

∴ AB : BC = 2k+k : k = 3 : 1

We need only to consider vertical coordinates.

Let the base line be $y = 0$ , and $y_A = 4$ . Then, because of the line segments of which they are midpoints, $y_F = 2$ , and hence $y_B = 1$ . Since $ABC$ is a straight line, we therefore know that $AB : BC = \boxed{3 : 1}$ .

Relevant wiki: Menelaus' Theorem

Applying Menelaus Theorem to the triangle $EFD$ and transversal $ABC$ , we get that

$\frac{ EB}{BF} \times \frac{FA}{AD} \times \frac{ DC}{CE} = 1$

Hence, we conclude that $\frac{ DC}{CE} = \frac{AD}{FA} \times \frac{BF}{EB} = \frac{2}{1} \times \frac{1}{1} = \frac{2}{1}$ . This tells us that $\frac{ED}{CE} = 1 + \frac{DC}{CE} = 3$ .

Applying Menelaus Theorem to the triangle $ACD$ and transversal $EBF$ , we get that

$\frac{ AB}{BC} \times \frac{CE}{ED} \times \frac{ DF}{FA} = 1$

Hence we concluce that $\frac{AB}{BC} = \frac{FA}{DF} \times \frac{ED}{CE} = \frac{1}{1} \times \frac{3}{1} = 3$ .