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

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Math Expert
Joined: 02 Sep 2009
Posts: 46305
In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z = [#permalink]

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17 Aug 2017, 23:43
00:00

Difficulty:

75% (hard)

Question Stats:

72% (01:53) correct 28% (02:54) wrong based on 122 sessions

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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

Attachment:

2017-08-18_1027_001.png [ 14.35 KiB | Viewed 2851 times ]

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Joined: 18 Aug 2016
Posts: 634
GMAT 1: 630 Q47 V29
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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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18 Aug 2017, 01:19
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

Attachment:
2017-08-18_1027_001.png

angle BAC = angle BCA = z
angle BCD = angle DCA = z/2
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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18 Aug 2017, 08:33
1
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

Attachment:
The attachment 2017-08-18_1027_001.png is no longer available

Attachment:

ABCxyztriangle.png [ 24.49 KiB | Viewed 2205 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 $$z = 4y$$, $$z = 80$$

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Intern
Joined: 17 Apr 2017
Posts: 9
Location: India
WE: Analyst (Consulting)
In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z = [#permalink]

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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
Intern
Joined: 08 Jun 2015
Posts: 21
Location: United States
Concentration: Finance, Economics
WE: Engineering (Other)
Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z = [#permalink]

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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.
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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