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In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =
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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 Attachment:
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Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =
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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: 20170818_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 = 180x+y+z/2 xyz/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 =
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18 Aug 2017, 08:33
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 20170818_1027_001.png is no longer available Attachment:
ABCxyztriangle.png [ 24.49 KiB  Viewed 4532 times ]
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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In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =
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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 = 180x
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, \(x/3+180x+z/2=180\) => \(z/2 =2x/3\) => \(x =3z/4\) (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 =
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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.



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Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =
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26 Nov 2019, 18:08
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: 20170818_1027_001.png 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 1Now 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 2Now that we have 2 unique equations, we can isolate Z and solve it. Z = 80



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Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =
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28 Nov 2019, 17:43
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: 20170818_1027_001.png 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
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Re: In ∆ ABC above, AB = BC, and CD bisects angle C. If y = x/3, then z =
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28 Nov 2019, 17:43






