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
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: .$\displaystyle \angle1\,=\,\angle2$