WILL GIVE BRAINLIEST!
2017 AMC 12A Problem 24
Quadrilateral [tex]ABCD[/tex] is inscribed in circle [tex]O[/tex] and has side lengths [tex]AB=3, BC=2, CD=6[/tex], and [tex]DA=8[/tex]. Let [tex]X[/tex] and [tex]Y[/tex] be points on [tex]\overline{BD}[/tex] such that [tex]\frac{DX}{BD} = \frac{1}{4}[/tex] and [tex]\frac{BY}{BD} = \frac{11}{36}[/tex]. Let [tex]E[/tex] be the intersection of line [tex]AX[/tex] and the line through [tex]Y[/tex] parallel to [tex]\overline{AD}[/tex]. Let [tex]F[/tex] be the intersection of line [tex]CX[/tex] and the line through [tex]E[/tex] parallel to [tex]\overline{AC}[/tex]. Let [tex]G[/tex] be the point on circle [tex]O[/tex] other than [tex]C[/tex] that lies on line [tex]CX[/tex]. What is [tex]XF\cdot XG[/tex]?
SHOW ALL WORK!

Respuesta :

Answer:

  XF·XG = 17

Step-by-step explanation:

Drawing a cyclic quadrilateral from side lengths is interesting enough. I have no idea how to compute the desired value, but my geometry program says it is 17.

In the attached, XF is line segment m, approximately 7.63 in length. XG is line segment n, approximately 2.23 in length. The product is the number "o", shown at the end of the red arrow.

Ver imagen sqdancefan