Question

dec 3 the student will be able to find side and angle measures of triangles
ec ⊥ bisector of ab and ∠aeb
solve for x, y, z
triangle with vertices a, e, b; c is on ab. ae: 10z + 10, ec: 5x + 10, eb: 8x + 3y, ab: ac (6y - 10) and cb (4y + 10), eb: 8z + 30
x blank y blank z blank
blank ac= blank ae= blank

Explanation:

Step1: Solve for y (midpoint property)

Since EC is the perpendicular bisector of AB, AC = BC:
$6y - 10 = 4y + 10$
$2y = 20$
$y = 10$

Step2: Solve for x (angle bisector property)

EC bisects ∠AEB, so AE = BE:
$5x + 10 = 8x + 25$
$-3x = 15$
$x = -5$

Step3: Solve for z (congruent sides)

AE = BE = $8z + 30$, and AE = $5x + 10$. Substitute x=-5:
$5(-5) + 10 = 8z + 30$
$-25 + 10 = 8z + 30$
$-15 = 8z + 30$
$8z = -45$
$z = -\frac{45}{8}$

Answer:

$x = -5$, $y = 10$, $z = -\frac{45}{8}$
AC = $6(10) - 10 = 50$
AE = $5(-5) + 10 = -15$ (Note: Length can't be negative, possible typo in problem)