Challenging Geometry Puzzle: Problem 1565. Share your solution by posting it in the comment box provided.

Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Extent BD=DG. Draw CH┴DG (H on DG)

ReplyDeleteTrABD, TrCGD are congruence (SAS) so GC=AB=CF,<ABD=<G

TrEAB,TrHCG are congruence (Right Triangles, CG=AB,<ABE=<G) so

BE=HG but HG=HF (TrCGH=TrCFH) so BE=HF

BE=HF → BF+EF=EF+EH → BF=EH=5+5=10 (EH=2ED)

On the line BD take a point G symmetric to B relative to D, so that DG = BD. Then triangle DGC is congruent to ABD (AD = DC, BD = DG, angle <BDA = <CDG, SAS case). Project point C to the line BDG, and call it H. As AD = DC, <EDA = <CDH,

ReplyDeleteso DH = ED = 5.

Also HG = BE, because ABE and CGH are right triangles, angle <ABE = <CGH and CG = AB.

Now, CF = AB (by hypothesis) and CG = AB (congruence of triangles ABD and DGC as seen before), implies that CF = CG and triangle FCG is isosceles. Then H is the middle point of FG, so that FH = HG, and as we saw HG = BE, and then FH = BE.

But,

FH= FE + ED + DH = FE + 5 + 5 = FE + 10;

BE = BF + FE,

So FH = BE implies FE + 10 = BF + FE, and cancelling FE we obtain BF = 10.

Joaquim Maia

Rio de Janeiro

Brazil

Extend ED to X such that D is the midpoint of EX

ReplyDeleteThen AECX is a parallelogram and AE = CX

Hence Triangles ABE & CFX are congruent (Hypotenuse, side, right angle)

So BE = FX, hence BF = EX = 10

Sumith Peiris

Moratuwa

Sri Lanka