1. ## Intersecting lines

In the figure below, transversal $r$ crosses both $p$ and $q$, and angle $a$ and $b$ are measures of the indicated angles, both between $0$ and $180$. Lines $p$ and $q$ will cross somewhere to the left of transversal $r$ (that is, on the side opposite the indicated angles).
Question: Why is $a + b > 180$ a true relationship between $a$ and $b$ for all possible positions of transversal $r$?

2. Hello, dannyc!

$\text}In the figure below, transversal }r\text{ crosses both }p\text{ and }q,$
$\text{and angles }a\text{ and }b\text{ are measures of the indicated angles,}$
$\text{both between }0^o\text{ and }180^o.$

$\text{Lines }p\text{ and }q\text{ intersect somewhere to the left of transversal }r.$

$\text{Why is }a + b \,>\, 180^o\text{ true for all possible positions of transversal }r\,?$

Code:
                                   r
/       p
K /    *
o
*     / a
*           /
*                 /
J o                       /
*               /
*       / b
o
/ L    *
*
q

Let: . $\begin{Bmatrix} p \cap q &=& J \\
p \cap r &=& K \\ q \cap r &=& L \end{Bmatrix}$

We have: . $\begin{Bmatrix}\angle JKL \,=\,180^o - a \\ \angle JLK \,=\,180^o - a \end{Bmatrix}$

Hence: . $\angle J \;=\; 180^o - (180^o - a) - (180^o - b) \;=\;(a+b) - 180^o$

Since $\angle J$ is a positive angle: . $(a+b)-180^o \:>\:0^o$

Therefore: . $a + b \:>\:180^o$

3. Originally Posted by Soroban
Since $\angle J$ is a positive angle: . $(a+b)-180^o \:>\:0^o$

Thanks Soroban! I think I got it now! (Btw, JLK = 180 - b above)