## Thursday, July 18, 2024

### Geometry Problem 1565: Find the Length of BF in Triangle ABC

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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1. Georgios Kousinioris, Gastouni GreeceJuly 21, 2024 at 1:11 AM

Extent BD=DG. Draw CH┴DG (H on DG)
TrABD, 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)

2. 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,
so 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

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

Then 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