Here is one way.

Let angle CEF and angle ACB be theta each.

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In triangle ADC:

Let x = side DC.

By Law of Sines,

(AD = 15)/sin(theta) = x/sin(angle CAD)

Cross multiply,

x*sin(theta) = 15*sin(angle CAD)

sin(theta) = [15*sin(angle CAD)]/x ----(i)

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In triangle EFC:

FD = 15-8 = 7 -----why?

EF = BF = 8 ----that means triangle BFE is isosceles.

AB is parallel to EB; BD is parallel to BF; and AD is parallel to EF. Therefore, triangle BDA is similar to triangle BFE.

Hence, triangle BDA is also isosceles, and since AD=15, then BD=15 also.

Since BF=8, then FD = BD-BF = 15-8 = 7 ----***

So, by Law of Sines again,

(FC)/sin(theta) = (EF)/sin(angle FCE)

Substitutions,

(7+x)/sin(theta) = 8/sin(angle FCE)

Cross multiply,

8*sin(theta) = (7+x)*sin(angle FCE)

sin(theta) = [(7+x)sin(angle FCE)]/8 ----(ii)

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sin(theta) from (i) = sin(theta) from (ii),

[15sin(angle CAD)]/x = [(7+x)sin(angle FCE)]/8

Cross multiply,

x[(7+x)sin(angle FCE)] = 8[15sin(angle CAD)]

(7x +x^2)sin(angle FCE) = 120sin(angle CAD) ----(iii)

You have noticed maybe that I am about to show that (angle FCE) = (angle CAD).

In the figure, in triangles EFC and ADC,

Since EF and AD are parallel, then (angle EFC) = (angle ADC).

Given: (angle CEF) = (angle ACB, same as angle ACD)

That means: (two angles of triangle EFC) = (two angles of triangle ADC).

That means further that their third angles are equal also.

Hence, (angle FCE) = (angle CAD) ----***

Substitute that into (iii),

(7x +x^2)sin(angle FCE) = 120sin(angle CAD) ----(iii)

(7x +x^2)sin(angle CAD) = 120sin(angle CAD)

7x +x^2 = 120

x^2 +7x -120 = 0 ----(iv)

Factor that, or use the quadratic formula, and you will get x = 8.

Therefore,

BC = BF +FD +x

BC = 8 +7 +8

BC = 23 ----answer.