In a triangle ABC, D and F are points on BC, and E is a point on AB such that AD is parallel to EF, AngleCEF = AngleACB, AD = 15m, EF = 8m and BF = 8m. Find the length of BC.
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.