In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG.

FG=10

Angle AFB  =  Angle DFE

Angle DBA  = Angle FDE

Therefore, by AA congruency  Triangle DEF  is similar to Triangle BAF

But  DE  is  2/3  of DC.... and since AB  = DC, then  DE  is also 2/3 of AB

Then all  parts of triangle DEF  are 2/3  of  their corrresponding parts in triangle BAF

Thus....the altitude of triangle  DEF  is 2/3  of the altitude of triangle BAF

But the altitude of triangle DEF  = CG since they are parallel

And for the same reason, the altitude of triangle BAF   = BG

Thus  CG   =  2/3  of BG.....so ...  there are 5 parts of BC......and BG  is 3 of these

So  BG  =  3/5   of BC

And triangle BGF  is similar to  triangle BCD

But  BG  =  3/5  of BC....so  GF  =  3/5 of CD

So    GF  = FG  =   (3/5)  20    =   12

In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG.

\(\text{Let $DC = 20$ } \\ \text{Let $DE = \dfrac23\cdot 20$ } \\ \text{Let $EC = \dfrac13\cdot 20$ } \\ \text{Let $FG = x + EC $ or $x=FG-EC $ }\\ \text{Let $BG = p $ } \\ \text{Let $GC = q $ } \)

1. intercept theorem

\(\begin{array}{|rcll|} \hline \dfrac{q}{x} &=& \dfrac{p}{DE-x} \quad & \quad DE= \dfrac23\cdot 20 \\\\ \dfrac{q}{x} &=& \dfrac{p}{\dfrac23\cdot 20-x} \\\\ \mathbf{\dfrac{p}{q}} & \mathbf{=} & \mathbf{\dfrac{40-3x}{3x} \qquad (1)} \\ \hline \end{array} \)

2. intercept theorem

\(\begin{array}{|rcll|} \hline \dfrac{q}{DE-x} &=& \dfrac{p}{x+EC} \quad & \quad DE= \dfrac23\cdot 20 \qquad EC=\dfrac13\cdot 20 \\\\ \dfrac{q}{ \dfrac23\cdot 20-x} &=& \dfrac{p}{x+\dfrac13\cdot 20} \\\\ \mathbf{\dfrac{p}{q}} & \mathbf{=} & \mathbf{\dfrac{3x+20}{40-3x} \qquad (2)} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)=(2): & \mathbf{\dfrac{p}{q}} = \mathbf{\dfrac{40-3x}{3x} }&\mathbf{=}& \mathbf{\dfrac{3x+20}{40-3x}} \\\\ & \dfrac{40-3x}{3x} & = & \dfrac{3x+20}{40-3x} \\\\ & (40-3x)^2 & = & 3x(3x+20) \\\\ & 1600-240x+\not{9x^2} & = & \not{9x^2} +60x \\\\ & 1600-240x & = & 60x \\\\ & 300x &=& 1600 \quad & | \quad : 300 \\\\ & x &=& \dfrac{1600}{300} \\\\ & x &=& \dfrac{16}{3} \\\\ & FG &=& x + EC \qquad EC=\dfrac{20}{3} \\\\ & &=& \dfrac{16}{3} + \dfrac{20}{3} \\\\ & &=& \dfrac{36}{3} \\\\ & \mathbf{FG} &\mathbf{=} & \mathbf{12} \\ \hline \end{array}\)

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