1. ## Triangles

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.

2. Here is one way.

Let angle CEF and angle ACB be theta each.

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Let x = side DC.
By Law of Sines,
Cross multiply,

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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),
Cross multiply,
(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 = 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

3. Thanks a lot, that is excellent, I could not find a way to prove that triangle AEC was congruent to triangle CEF.

Although i did get a ratio: where y = CD

Then (y+7)/15 = y/8

y = 8m

Thanks once again!

4. One more thing ticbol, I love your style in working out and providing very throuogh answers. Also the writing of 'answer' at the end is also a fond habit of mine in all exams.

Once again keep up the fab work.