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

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

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New post 17 Aug 2017, 23:43
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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

[Reveal] Spoiler:
Attachment:
2017-08-18_1027_001.png
2017-08-18_1027_001.png [ 14.35 KiB | Viewed 1300 times ]
[Reveal] Spoiler: OA

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Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z = [#permalink]

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New post 18 Aug 2017, 01:19
Bunuel wrote:
Image
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

[Reveal] Spoiler:
Attachment:
2017-08-18_1027_001.png


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
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Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z = [#permalink]

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New post 18 Aug 2017, 08:33
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Bunuel wrote:
Image
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

[Reveal] Spoiler:
Attachment:
The attachment 2017-08-18_1027_001.png is no longer available

Attachment:
ABCxyztriangle.png
ABCxyztriangle.png [ 24.49 KiB | Viewed 958 times ]

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 above, AB = BC, and CD bisects angle C. If y = x/3, then z = [#permalink]

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New post 19 Aug 2017, 03:42
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

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Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z = [#permalink]

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New post 19 Aug 2017, 05:55
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.

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Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =   [#permalink] 19 Aug 2017, 05:55
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In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =

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