# Geometry Problem

## Main Question or Discussion Point

In $$\bigtriangleup ABC , AB = BC$$ and $$CD$$ bisects angle C.
If $$y = \frac{1}{3}x$$ then z = ....

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Ok. I've tried to make a bunch of substitutions for each of the angles and I can't seem to solve any for z.

I have found these two equations:

$$y + z + c = 180$$

and

$$x + z + \frac{c}{2} = 180$$

I know I can make substitutions/variants of these, but I still am stuck.
Any help is appreciated.

Jameson

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cepheid
Staff Emeritus
Gold Member
Notice that angle CDB = 180 - x. So:

(180 - x) + y + C/2 = 180 (from the upper triangle).

(180 - x) + x/3 + C/2 = 180

C/2 = 2x/3

Can you proceed from here?

I got the same thing. $$\frac{c}{2} = \frac{2x}{3}$$ but unfortunately I still can't seem to do the right substitution.

Substituting (2x)/3 for c/2 I get the equation:

$$x + \frac{2x}{3} + z = 180$$

This still leaves two variables. I just can't see it for some reason.

This is from a published SAT book from College Board, so I think there is a solution.

cepheid
Staff Emeritus
Gold Member
I, too, assumed that you were meant to solve for z in terms of x, since y was given in terms of x. I don't see what else you can do with it...

cepheid said:
I, too, assumed that you were meant to solve for z in terms of x, since y was given in terms of x. I don't see what else you can do with it...
no no, see how AB = BC, That's a very important clue. We know triangle ABC is an isoceles triangle with angle A = angle C. You can come up with an equation to relate x directly to z and substitute it back into one of the equations to solve for z.

Angle C = 4/3 x = Angle A = z

you should be able to take it from here.

honestrosewater
Gold Member
Eh, doesn't 2z + y = 180? So knowing y = x/3 you can solve for z in terms of x.

Plus, if it's the CB's "10 Real SAT's", it has the answers. And any other SAT prep book should have the answers too.

Last edited:
honestrosewater said:
Eh, doesn't 2z + y = 180? So knowing y = x/3 you can solve for z in terms of x.

Plus, if it's the CB's "10 Real SAT's", it has the answers. And any other SAT prep book should have the answers too.
ok, that equation is essential to solve for z. However, an actual value can be determined here. It's not impossible. I solved it and i got z = 80 degrees. Let me know if you want me to show my work.

apchemstudent... 80 degrees is the correct answer. I would be most gracious if you could show your work.

Jameson

Jameson said:
apchemstudent... 80 degrees is the correct answer. I would be most gracious if you could show your work.

Jameson
ok, we set angle BDC = 180 - x, so that we can solve for Angle BCD

BCD + DBC + CDB = 180 degrees
BCD = 180 -1/3 x - 180 + x
BCD = 2/3 x

BCD = 1/2 BCA so BCA = 4/3 x

Again we know that it is an isoceles triangle with angle BCA = BAC
z = 4/3 x

so BAC + ABC + BCA = 180
z + 1/3 x + 4/3 x = 180
3x = 180
x = 60 <---- you could've substituted x in terms of z and solved for z but i find it easier this way.

now z = 4/3 (60) = 80 degrees.