Circle Tangent Lengths
Given circle Γ , point A is chosen outside of Γ . Tangents A B and A C to Γ are drawn, such that B and C lie on the circumference of Γ . K is a point on the circumference of Γ contained within A B C . Let D , E and F be the foot of the perpendiculars from K to B C , A C and A B , respectively. If K F = 2 0 and K E = 2 8 , what is K D 2 ?
The answer is 560.
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2 solutions
From the alternate segment theorem, we have ∠ F B K = ∠ K C D and ∠ K F B = ∠ K D C = 9 0 ∘ , thus triangles F B K and K D C are similar by angle-angle-angle, so K D K F = K C K B .
Similarly, triangles E C K and K D B are similar, so K D K E = K B K C . Multiplying the two equations we have K D K F ⋅ K D K E = K B K C ⋅ K C K B = 1 .
Hence K D 2 = K E ⋅ K F = 2 8 ⋅ 2 0 = 5 6 0 .
What you call the alternate segment theorem,I know it as the tangent secant theorem,otherwise even my approach was the same as yours.
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Yup, those are the names given that that theorem about angles in a circle.
We have ∠ D B K = ∠ C B K = ∠ A C K = ∠ E C K , so triangles D B K and E C K are similar. Likewise, we have ∠ D C K = ∠ B C K = ∠ A B K = ∠ F B K , so triangles D C K and F B K are also similar. Thus, D K / K E = B K / C K and D K / K F = C K / B K . Multiplying these two equations we get K D 2 = K E × K F . Thus, K D 2 = 2 0 × 2 8 = 5 6 0 .