A geometry problem by A Former Brilliant Member

Geometry Level 3

In the figure above, A B C D ABCD is a parallelogram. Point E E lies on B C BC . A E AE cuts diagonal B D BD at G , G, and is extended to meet the extension of D C DC at F F . If A G = 6 AG=6 and G E = 4 GE=4 , find E F EF .


The answer is 5.

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1 solution

F D G \triangle FDG and A B G \triangle ABG are similar triangles,

\therefore G F G A = G D G B \dfrac{GF}{GA}=\dfrac{GD}{GB}

B G E \triangle BGE and A G D \triangle AGD are similar triangles,

\therefore G D G B = G A G E \dfrac{GD}{GB}=\dfrac{GA}{GE}

We equate G D G B = G D G B \dfrac{GD}{GB}=\dfrac{GD}{GB} , we have

G F G A = G A G E \dfrac{GF}{GA}=\dfrac{GA}{GE}

Substituting, we get

4 + E F 6 = 6 4 \dfrac{4+EF}{6}=\dfrac{6}{4}

4 + E F = 36 4 4+EF=\dfrac{36}{4}

4 + E F = 9 4+EF=9

E F = 9 4 EF=9-4

E F = 5 \boxed{EF=5} answer \boxed{\text{answer}}

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