Similar triangles

Geometry Level 2

In the isosceles triangle shown above, A D AD is an altitude. Given that A B = 3 2 B C , A F = 4 F D AB=\dfrac{3}{2}BC, AF=4FD and F D = 1 , FD=1, find F E FE .

4 3 \dfrac{4}{3} 3 2 \dfrac{3}{2} 5 4 \dfrac{5}{4} 3 4 \dfrac{3}{4}

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1 solution

Note that △ A E F ∼ △ A D C \triangle AEF \sim \triangle ADC . Then A F A C = F E D C = F E 1 2 B C = 2 F E B C \dfrac{AF}{AC}=\dfrac{FE}{DC}=\dfrac{FE}{\frac{1}{2}BC}=\dfrac{2FE}{BC}

Substituting, we get

4 A C = 2 F E B C \dfrac{4}{AC}=\dfrac{2FE}{BC}

Cross-multiplying and substituting, we get

F E = 4 B C 2 A C = 2 B C A C = 2 ( 2 3 ) A B A B = FE=\dfrac{4BC}{2AC}=\dfrac{2BC}{AC}=\dfrac{2(\frac{2}{3})AB}{AB}= 4 3 \boxed{\dfrac{4}{3}}

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