# Math Help - Help: two column proof involving triangles/properties of parallel lines!

1. ## Help: two column proof involving triangles/properties of parallel lines!

I have no idea how to start this, let alone finish it. The worst part is I have to solve it by the end of tonight! Would one of you guys please help me?
Oh yeah: can I assume triangle BST is an equilateral triangle?

Solve using a two column proof:
Given: Line segment AS is parallel to line segment BT;
The measure of angle 4 is equal to the measure of angle 5
Prove: Ray SA bisects angle BSR

Thank you!

2. By corresponding angles we know that 1 & 4 are congruent.
By alternant interior angles we know that 2 & 5 are congruent.

3. Originally Posted by frostbite
I have no idea how to start this, let alone finish it. The worst part is I have to solve it by the end of tonight! Would one of you guys please help me?
Oh yeah: can I assume triangle BST is an equilateral triangle?

Solve using a two column proof:
Given: Line segment AS is parallel to line segment BT;
The measure of angle 4 is equal to the measure of angle 5
Prove: Ray SA bisects angle BSR

Thank you!
No, BST is an isoceles triangle, but not necessarily equiliteral.

Set angle 5 = x
Angle 5 = Angle 4 (Given)
Angle 4 = x

Angle 3 = 180 - 2x (interior angles of a triangle)

Angle 2 = Angle 5 = x (Parallel lines, alternant interior angles)

Angle 1 = 180 - (Angle 2 + Angle 3)
Angle 1 = 180 - (x + (180 - 2x))
Angle 1 = x

Since angle 1 = angle 2, the line bisects the angle

4. Hello, frostbite!

Can I assume triangle BST is an equilateral triangle? . . . . NO!

Solve using a two-column proof.

Given: $AS \parallel BT;\;\angle 4 \,=\,\angle 5$

Prove: $SA$ bisects $\angle BSR$

I'll let you supply the reasons.

We see that: . $\angle BSR \:=\:\angle1 + \angle2$ .[1]

$\angle BSR$ is an exterior angle of $\Delta BST$.
. . Hence: . $\angle BSR \:=\:\angle 4 + \angle 5$
Since $\angle4 = \angle5$, we have: . $\angle BSR \:=\:2\cdot\angle4$ .[2]

Equate [1] and [2]: . $\angle 1 + \angle 2 \;=\;2\cdot\angle 4$ .[3]

Since $\angle1 = \angle4$ . (alt-int. angles),
. . [3] becomes: . $\angle1 + \angle 2 \;=\;2\cdot\angle1$

and we have: . $\angle 2 \:=\:\angle 1$

Therefore, $SA$ bisects $\angle BSR$.

5. Wow! I didn't expect so many replies so fast...
Thank you for the help!

6. You can expect 5 replies a minute at this forum.

7. That's exactly why this is the only forum that I've ever posted on more than twice. All thanks to a few dozen key people. Good job!