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

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

Click the figure below to view more details of problem 1289.

## Monday, November 28, 2016

### Geometry Problem 1289 Triangle, Cevian, Concurrency, Measurement, Metric Relations

Labels:
cevian,
concurrent,
measurement,
metric relations,
triangle

Subscribe to:
Post Comments (Atom)

Problem 1289

ReplyDeleteFrom Ceva’s theorem BF/FA.AE/EC.CD/DB=1, from Menelaus

Theorem BF/FA.AG/GC.CD/DB=1 then BF/FA.AE/EC.CD/DB=BF/FA.AG/GC.CD/DB or AE/EC=AG/GC or

5/3=(8+x)/x or x=12.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

Observe that AFDCBG is a complete quadrilateral with diagonals AD and FC met at O

ReplyDeleteSo E and G are the harmonic conjugate of A and C or (ACEG)=-1

and EC/EA=GC/GA=3/5= x/(x+8) => x= 12

https://en.wikipedia.org/wiki/Projective_harmonic_conjugate A,E,C,G form a system of Harmonic Conjugate points therefore 3/5=(x/x+8) -->x=12

ReplyDelete