1. ## Question about triangles and their exterior angle

I got this question wrong in my GRE math prep, and I have no idea why.

Triangle ABC is an isosceles triangle with AB = BC. The side AC is extended to point D, so that BCD forms an exterior angle. (See attached picture)

Which of the following can be concluded?

Among a few other answers, I circled: Measure of angle BCD is greater than measure of angle BCA.

This was not one of the answers in the answer key, and the explanation said that the relative measures of angle BCD and BCA could not be determined.

I feel that that is somehow wrong because triangle ABC is an isosceles.

I played with it for a long time, trying to prove it, and this is what I came up with, but I'm not certain at all. (Also I don't know how to do the angle sign, so I'm just going to use an < for now, and I've used a picture with the angles labeled with numbers to make it easier to identify).

<1 = <3 because triangle ABC is isosceles, with AB=BC
<1 + <2 +<3 = 180
<3 + <4 = 180

If <1=<3, then 2 times <3 + <2 = 180.

2 times <3 + <2 = 180 = <3 + <4
<3 + <2 = <4

So, <4 must be greater than <3 unless <2 =0

I don't know if this is correct or not, can anyone explain it to me in another way?

2. ## Re: Question about triangles and their exterior angle

$\angle 1 + \angle 2 + \angle 3 = 180^\circ$. Since $\angle 1 = \angle 3$ it can easily be shown that

$\angle 3 < 90^{\circ}$. It follows that $\angle 4 > 90^{\circ}$, and $\angle 4 > \angle 3$.

3. ## Re: Question about triangles and their exterior angle

Oh, I hadn't even thought of that. Of course, because otherwise <3 + <1 would be 180.

So, that is something I can conclude after all? Silly prep book

4. ## Re: Question about triangles and their exterior angle

Yep. A lot of prep books are badly flawed.

5. ## Re: Question about triangles and their exterior angle

Originally Posted by Blueberry
I got this question wrong in my GRE math prep, and I have no idea why.

Triangle ABC is an isosceles triangle with AB = BC. The side AC is extended to point D, so that BCD forms an exterior angle. (See attached picture)

Which of the following can be concluded?

Among a few other answers, I circled: Measure of angle BCD is greater than measure of angle BCA.

This was not one of the answers in the answer key, and the explanation said that the relative measures of angle BCD and BCA could not be determined.

I feel that that is somehow wrong because triangle ABC is an isosceles.

I played with it for a long time, trying to prove it, and this is what I came up with, but I'm not certain at all. (Also I don't know how to do the angle sign, so I'm just going to use an < for now, and I've used a picture with the angles labeled with numbers to make it easier to identify).

<1 = <3 because triangle ABC is isosceles, with AB=BC
<1 + <2 +<3 = 180
<3 + <4 = 180

If <1=<3, then 2 times <3 + <2 = 180.

2 times <3 + <2 = 180 = <3 + <4
<3 + <2 = <4

So, <4 must be greater than <3 unless <2 =0

I don't know if this is correct or not, can anyone explain it to me in another way?
Since

$\angle1 + \angle2 + \angle3 = 180^\circ$ ....... and

$\angle3 + \angle4 = 180^\circ$

follows: $\angle4 = \angle1 + \angle2$

6. ## Re: Question about triangles and their exterior angle

Hello, Blueberry!

I got this question wrong in my GRE math prep, and I have no idea why.

Triangle ABC is an isosceles triangle with AB = BC.
The side AC is extended to point D, so that BCD forms an exterior angle. (See attached picture)

Code:
            B
o
* *
* 2 *
*     *
*       *
* 1     3 * 4
A o * * * * * o * * * o D
C
Which of the following can be concluded?

Among a few other answers, I circled: Measure of / BCD is greater than measure of / BCA.
. . You are correct!

The explanation said that the relative measures of / BCD and BCA could not be determined.
. . This is wrong!

I feel that that is somehow wrong because triangle ABC is an isosceles. . Right!

Putting it simply: $\angle 3$ must be acute; therefore, $\angle 4$ is obtuse.

Here is another proof . . .

We know that: . $\angle 1 + \angle 2 + \angle 3 \:=\:180^o$

Since $AB = BC,\:\angle 1 = \angle 3.$

So we have: . $2\angle 3 + \angle 2 \:=\:180^o$

Since $\angle 2 > 0^o,\,\text{ then: }\,2\angle 3 \:<\:180^o$

Therefore: . $\angle 3 \:<\:90^o\:\text{ . . . }\:\angle 3\text{ is acute.}$

Even more obvious: $\text{If }\angle 3 \ge 90^o,\:there\;is\;no\;triangle.$