In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =

(A) 40
(B) 60
(C) 64
(D) 72
(E) 80


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Derivation of figure's annotations:

AB = BC: ∠ A = ∠ C
CD bisects ∠ C: two angles formed by the bisector are equal
∠ A = ∠ C, and ∠ A = \(z\). So ∠ C = \(z\)
Angles at ∠ C formed by bisector = \(z/2\)

Isolate one variable:

1) Given: \(y = \frac{x}{3}\). \(x = 3y\)

\(x\) shows up in the figure only once; replace it with \(3y\) in figure 2

∠ BDC and 3y are on a straight line. ∠ BDC = \((180 - 3y)\)

2) All three triangles' angles measures now are expressed in terms of y and z. Find one variable in terms of the other.

\(∆ ABC: 2z + y = 180\)

\(∆ ADC: z + \frac{z}{2} + 3y = 180\)

\(∆ BDC: y + (180 - 3y) + \frac{z}{2} = 180\)

3) Use ∆ BDC first: with 180 on both sides of the equation, RHS will = 0. That's a good start.

\(y + (180 - 3y) + \frac{z}{2} = 180\)

\(y + 180 - 3y + \frac{z}{2} = 180\)

\(-2y + \frac{z}{2}= 0\)

\(\frac{z}{2} = 2y\),

\(z = 4y\)

4) Find \(y\), where \(z = 4y\). ∆ ABC is simplest for substitution

\(2z + y = 180\)

\(2(4y) + y = 180\)

\(9y = 180\)

\(y = 20\)

5) If \(y = 20\), from \(z = 4y\), \(z = 80\)

Answer E
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In ∆ ABC z+y+c=180 or 2z+y=180 as AB=BC so z=c (Iso. triangle). So we can write it as 12z+2x=1080.....(1)

In ∆ ADC, z+x+c/2 = 180 or 3z+2x=360......(2)

Solving 1 & 2 we get z=80, So IMO "E" is correct answer.