A geometry problem by A Former Brilliant Member

Geometry Level pending

In the figure above, E D F EDF is a tangent to circle O O . Triangle A B C ABC is inscribed in the circle. A B AB is extended to point E E and A C AC is extended to point F F . If A B = 4 AB=4 , B E = 8 BE=8 and A C = 6 AC=6 , find C F CF .

Note: The figure is not drawn to scale.


The answer is 2.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Draw D C DC and B D BD .

In right △ A D E : \triangle ADE: A E A D = A D A B \dfrac{AE}{AD}=\dfrac{AD}{AB} . Thus, ( A D ) 2 = ( A E ) ( A B ) (AD)^2=(AE)(AB) .

In right △ A D F : \triangle ADF: A F A D = A D A C \dfrac{AF}{AD}=\dfrac{AD}{AC} . Thus, ( A D ) 2 = ( A F ) ( A C ) (AD)^2=(AF)(AC) .

( A D ) 2 = ( A D ) 2 (AD)^2=(AD)^2

( A E ) ( A B ) = ( A F ) ( A C ) (AE)(AB)=(AF)(AC)

Substituting, we get

( 12 ) ( 4 ) = ( 6 + C F ) ( 6 ) (12)(4)=(6+CF)(6)

8 = 6 + C F 8=6+CF

C F = 2 \boxed{CF=2}

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