Bunuel wrote:
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 = ∠ CCD 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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