Geometry Problem

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

Click the figure below to see the complete problem 829.

## Friday, November 30, 2012

### Problem 829: Triangle, Circumcircle, Parallel, Chord, Side, Metric Relations

Labels:
chord,
circumcircle,
metric relations,
parallel,
triangle

Subscribe to:
Post Comments (Atom)

x=5sqrt(5/2)

ReplyDeleteDenote BF, BG by p & q resply. Use Stewart's Theorem to obtain: 125+x^2=6(p^2+5), 50+4x^2=6(q^2+8)and p/(p+5/p) =q/(q+8/q).

ReplyDeleteSolve simultaneously to get x=5√(2.5) --- same as Atharva

Since DE//AC so Arc(AD)=Arc(EC) => angle (ABF)=angle(CBG)

ReplyDeleteSo in triangle ABC ,line BF and BG are isogonal to each other.

Per Steiner’s theorem for isogonal lines BF and BG

(AF/FC).(AG/GC)=AB^2/BC^2

So 1/5 . 4/2 =25/x^2

So x=5.sqrt(5/2)