Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Monday, October 8, 2018

### Geometry Problem 1394: Three Circles and Three Pair of Common Internal Tangents, Concurrency

Labels:
center,
circle,
common tangent,
concurrency,
concurrent,
internal,
three circles

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circle A, radius is Ra;

ReplyDeleteconnect AB, ab go through D,

same for AC,BC; E on AC, F on BC;

so AE/EC= Ra/Rc, CF/FB=Rc/Rb; BD/DA=Rb/Ra;

so AE/EC * CF/FB * BD/DA =1,

so AF,BE,CD concurrent

By ceva's theorem: (AE)/(EC).(CF/CB).(BD/DA)=1 (Tr AET similar to tr ECP T,P tg points)

ReplyDelete(Ra/Rc).(Rc/Rb).(Rb/Ra)=1

Fie triunghiul format de ABC. Se arată ușor că punctele E , F și D se găsesc pe laturile AC, CB și AB.

ReplyDeleteNotând cu M punctul de tangență cu cercul A și cu N punctul de tangență comună cu cercul C, din asemănarea triunghiurilor AME și CNE ( AA) se obține raportul AE/EC = r_A/r_C si analog CF/FB = r_c/r_B respectiv BD/DA = r_B/r_A . De aici AE/EC ∙ CF/FB ∙ BD/DA= r_A/r_C ∙ r_A/r_C ∙ r_B/r_A = 1 și conform reciproca teoremei Ceva rezultă că dreptele AF, CD, BE sunt concurente.

Florin Popa , Comănești, România.

https://photos.app.goo.gl/bEcZvA4JTfREGsin9

ReplyDeleteLet rA, rB, rC are radius of circles A, B and C

Note that B,F,C are colinear

Triangle BB’F similar to CC’F ( case AA)

So BF/FC= BB’/CC’= rB/rC

Similarly we also have CE/AE= rC/rA and DA/DB= rA/rB

Multiply these expressions side by side and simplify it we get

BF/BC x EC/EA x DA/DB= 1

So AF, BE, CD are concurrent per Ceva’s theorem