Circle Tangent Lengths

Geometry Level 5

Given circle Γ \Gamma , point A A is chosen outside of Γ \Gamma . Tangents A B AB and A C AC to Γ \Gamma are drawn, such that B B and C C lie on the circumference of Γ \Gamma . K K is a point on the circumference of Γ \Gamma contained within A B C ABC . Let D , E D, E and F F be the foot of the perpendiculars from K K to B C , A C BC, AC and A B AB , respectively. If K F = 20 KF = 20 and K E = 28 KE = 28 , what is K D 2 KD^2 ?


The answer is 560.

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2 solutions

We have D B K = C B K = A C K = E C K \angle DBK = \angle CBK = \angle ACK = \angle ECK , so triangles D B K DBK and E C K ECK are similar. Likewise, we have D C K = B C K = A B K = F B K \angle DCK = \angle BCK = \angle ABK = \angle FBK , so triangles D C K DCK and F B K FBK are also similar. Thus, D K / K E = B K / C K DK/KE=BK/CK and D K / K F = C K / B K DK/KF = CK/BK . Multiplying these two equations we get K D 2 = K E × K F KD^2=KE \times KF . Thus, K D 2 = 20 × 28 = 560 KD^2=20\times 28 = 560 .

Calvin Lin Staff
May 13, 2014

From the alternate segment theorem, we have F B K = K C D \angle FBK = \angle KCD and K F B = K D C = 9 0 \angle KFB=\angle KDC = 90^\circ , thus triangles F B K FBK and K D C KDC are similar by angle-angle-angle, so K F K D = K B K C \frac {KF}{KD}=\frac {KB}{KC} .

Similarly, triangles E C K ECK and K D B KDB are similar, so K E K D = K C K B \frac {KE}{KD} = \frac {KC}{KB} . Multiplying the two equations we have K F K D K E K D = K C K B K B K C = 1 \frac{KF}{KD}\cdot \frac{KE}{KD} = \frac{KC}{KB}\cdot \frac{KB}{KC} = 1 .

Hence K D 2 = K E K F = 28 20 = 560 KD^2 = KE \cdot KF = 28 \cdot 20 = 560 .

What you call the alternate segment theorem,I know it as the tangent secant theorem,otherwise even my approach was the same as yours.

Indraneel Mukhopadhyaya - 5 years, 4 months ago

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Yup, those are the names given that that theorem about angles in a circle.

Calvin Lin Staff - 5 years, 3 months ago

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