6. I is the incenter of Triangle XYZ. A circle centered at I intersects the three sides of Triangle XYZ at the six points A,B C,D E,and F as shown below. We know that YZ=4, ZX=5, and XY=6=AB+CD+EF. Find XA.
Draw IX, IY, IZ
Draw IM perpendicular to YZ
Draw IN perpendicular to XZ
Draw IO perpendicular to XY
By AAS
Triangle IMZ congruent to Triangle INZ
Triangle INX concruent to Triangle IOX
Triangle IMY congruent to Triangle IOY
So.....
IM = IN = IO
But....chords equidistant from the center of a circle are themselves equal
So AB = CD = EF
So
AB + CD + EF = 6 and by substitution, we have that
AB + AB + AB = 6
3AB = 6
AB = 2 = CD = EF
So
AB/2 = CD/2 = EF/2
OY = MY
AB/2 + BY = CD/2 + CY
OX = NX
AB/2 + AX = EF/2 + FX
NX = MX
EF/2 + EZ = CD/2 + DZ
So......this implies that
CY = BY = a
AX = FX = b
EZ = DZ = c
So......we have this system
a + CD + c = 4 ⇒ a + c = 2 ⇒ a = 2 - c (1)
a + AB + b = 6 ⇒ a + b = 4 (2)
b + EF + c = 5 ⇒ c + b = 3 ⇒ b = 3 - c (3)
Sub (1) and (3) into (2)
2 - c + 3 - c = 4
5 - 2c = 4
-2c = -1
c = 1/2
b = XA = 3 - c = 3 - 1/2 = 5/2
