Geometry Problem 1512. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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https://photos.app.goo.gl/Rr1mPzhRSN29rhfb6

ReplyDeleteLet CF meet AB at G

Apply Ceva theorem for concurrent point F we have (EA/EC) x(DC/DB) x( GB/GA)= 1

Replace DC/DB= 1 and EA/EC=½ we get GB/GA= 2

Apply Van Aubel theorem we have FB/FE= GB/GA + DB/DC = 2+1= 3

So BF= 3GA= 18

https://photos.app.goo.gl/cDwW2cJUjC13tptQ9

ReplyDeleteIf G is the mid point of BE, from the mid point theorem,

ReplyDeleteGD = CE/2 = AE

Then Tr.s AEF & DFG are congruent ASA

So GF = EF = 6 and so GE = 6 + 6 = 12

Hence BF = BG + GF = GE + GF = 12 + 6 = 18

Sumith Peiris

Moratuwa

Sri Lanka

From Rimitti / Wahran / Algeria

ReplyDeleteApply Menelau's theorem to triangle EBC and secant AFD :

(BD/DC)*(CA/AE)*(EF/FB) = 1 with CA= 3AE and BD=DC ---> 3*EF/FB = 1---> FB =18