angle BAC = angle BCA = z
angle BCD = angle DCA = z/2
in triangle ADC
180 = x+z+z/2 = x+3z/2.....(i)
in triangle BDC
180 = 180-x+y+z/2
x-y-z/2=0...........................(ii)
in triangle ABC
180=z+z+y
180=2z+y..........................(iii)

we know y=x/3
x=3y
2y=z/2
4y=z

2z+y=180
2z+z/4=180
9z/4=180
z=80, y=20 and x=60
E
_________________

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
_________________

Given that AB=BC,
Therefore Angle C = Angle A = z
Also, CD is the angular bisector for angle C, so Angle BCD = Angle ACD = z/2
and y= x/3

As we know Angle ADC+ Angle BDC = 180 (Straight angle)
x+Angle BDC =180
Angle BDC = 180-x

Now that we have all angles
Solve for Z using sum of angles property in triangles ADC and BDC
In ADC, \(z+x+z/2 =180\)
=> \(x+3z/2 =180\) --(1)

In BDC, \(x/3+180-x+z/2=180\)
=> \(z/2 =2x/3\)
=> \(x =3z/4\) --(2)
use value of x from (2) in (1)
\(3z/4 + 3z/2 = 180\)
Therefore z=80

Ans: E

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.

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

First, we see that angle CDB is (180 - x), so angle BCD is:

180 - y - (180 - x) = x - y

Since y = x/3, we can substitute this for y, obtaining x - x/3 = 2x/3 for angle CDB

Since angle C is bisected, by CD, we know that angle ACD is also 2x/3.

Thus, angle C is 4x/3, which is also equal to z (because triangle ABC is isosceles).

So, finally, we have:

4x/3 + 4x/3 + x/3 = 180

9x/3 = 180

3x = 180

x = 60

Recall that 4x/3 equals angle z; thus z = 4(60)/3 = 4*20 = 80.

Answer: E
_________________

I have not done geometry in a while, so this took me about 10 minutes, but I got the right answer thinking it through and remembering some basic concepts. IMPORTANT: It is helpful to use x=3y instead of y=x/3! Kudos to all the correct solutions!

Let this be a lesson that one algebra mistake can ruin it all...Took a colossal 9 minutes for a very straightforward question. This is all algebra.

y + 2z = 180 (Since we know that AB = BC) --> x/y + 2z = 180

For the smaller triangle:

x + z + z/2 = 180 --> 2z + z + 2x = 360 --> 3z + 2z = 360 --> 2x = 360 - 3x --> x = 180 - 3z/2 --> x = 360 - 3z/2

y = x/3 = 1/3 x (360 - 3z/2) = 360 - 3z/2 (plug this into the equation above)

360 - 3z/6 + 2z = 180 --> z = 80

E.
_________________