A geometry problem by A Former Brilliant Member

Geometry Level pending

In the triangle shown above, B D BD is a median. Point E E is the midpoint of B D BD . Line C E CE extended meets A B AB at F F . If A B = 6 AB=6 , find F B FB .


The answer is 2.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Let F B = x FB=x and apply Menelaus' Theorem to △ A B D \triangle ABD with transversal C F CF :

A F F B × B E E D × D C C A = − 1 \dfrac{AF}{FB} \times \dfrac{BE}{ED} \times \dfrac{DC}{CA} = -1

which implies that,

A F x × 1 × ( − 1 2 ) = − 1 \dfrac{AF}{x} \times 1 \times \left(-\dfrac{1}{2} \right)=-1

A F = 2 x AF=2x , so A B = 3 x AB=3x , and it follows that

x = A B 3 = 6 3 = x=\dfrac{AB}{3}=\dfrac{6}{3}= 2 \boxed{2}

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