# Thread: Two-Column proof with triangles, theorems?

1. ## Two-Column proof with triangles, theorems?

Man I hate proofs.

Given: segment AS//BT, and <4 congruent to <5
Prove: <1 congruent to <2

I would really appreciate if anyone could show the steps and theorems you used to solve this (we all know how proofs work)

2. ## Re: Two-Column proof with triangles, theorems?

There's only one tricky bit.

Because AS || BT we know <5 =<2

let <4 = <5 = x

working in degrees

2x+<3 = 180 as the sum of interior angles of a triangle is 180 degrees

<1 + <2 + <3 = 180 as these angles make up a straight angle

subtract the 1st from the 2nd equation

<1 + <2 - 2x = 0

<1 + <2 = 2x

Now <5=<2 = x

<1 + x = 2x

<1 = x

<1 = x = <2

<1 = <2

3. ## Re: Two-Column proof with triangles, theorems?

Hello, funnyguy287!

I don't have a two-column proof.
I'll show you my steps.
I'll let you break it up into Statements and Reasons.

$\displaystyle \text{Given: }\:AS \parallel BT,\;\;\angle 4 = \angle 5$
$\displaystyle \text{Prove: }\:\angle 1 = \angle 2.$

Code:
                              B
*
*6*5*
*   *   *
A   *     *     *
*       *       *
*   *     *         *
*       * 2 *           *
*         1 * * 3         4 *
* * * * * * * * * * * * * * * * *
R               S               T

Since $\displaystyle AS \parallel BT,\;\angle1 = \angle4.$

In $\displaystyle \Delta SBT,\:\angle 3 + \angle 4 + \angle 5 \:=\:180^o \quad\Rightarrow\quad \angle 3 \:=\:180^o - 2\angle 4$

We have: .$\displaystyle \angle RST \:=\:180^o$

. . $\displaystyle \angle1 + \angle2 +\angle3 \:=\:180^o$

. . $\displaystyle \angle 4 + \angle 2 + (180^o-2\angle4) \:=\:180^o$

. . $\displaystyle \angle2 - \angle4 \:=\:0 \quad\Rightarrow\quad \angle 2 = \angle 4$

Since $\displaystyle \angle1 = \angle4,$ we have: .$\displaystyle \angle1\,=\,\angle2$

4. ## Re: Two-Column proof with triangles, theorems?

given BS=BT AS parallel to BT a for angle
1 a2=a4
2a1=a5
3a4=a5
a1=a4
a1=a2