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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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For these type of questions, I like to derive the equations so I can isolate the variable of choice, which is Z From the Information above, we know that AB = BC, which means that ∠BAC = ∠BCA ∠BAC = Z, then ∠BCA = Z
∠ABC +∠BAC +∠BCA =180, or Y+Z+Z=180 Equation 1
Now look at Triangle △ADC ∠ADC is equal to x, or 3y, and ∠DCA is equal to 0.5z because Line CD bisects the triangle, which means it divides ∠BCA by 2.
∠ADC+∠DAC+∠DCA = 180 3Y+Z+0.5Z=180 Equation 2
Now that we have 2 unique equations, we can isolate Z and solve it. Z = 80
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68% (02:52) correct 
