symmetry with respect to the x-axis and the y-axis Please help me prove that if a graph is symmetric with respect to the x-axis and to the y-axis, then it is symmetric with respect to the origin.
 A: A graph is symmetric about the $x$-axis if and only if whenever $(a,b)$ is in the graph, so is $(a,-b)$.
A graph is symmetric about the $y$-axis if and only if whenever $(a,b)$ is in the graph, so is $(-a,b)$.
A graph is symmetric about the origin if and only if whenever $(a,b)$ is in the graph, so is $(-a,-b)$.
Say you have a point $(a,b)$ on the graph. Can you show (say, in a couple of steps), that symmetry about $x$ and symmetry about $y$, together, imply that $(-a,-b)$ has to be in the graph as well?
A: Take a point $a$ in the first quadrant (without loss of generality). 
Draw a line $l_1$from the origin to $a$ and let $\theta$ be the angle formed by $l_1$ and the x axis.
Reflect $a$ about the $x$ axis and call that point $b$. 
Draw the line $l_2$ from the origin to $b$ and call $\beta$ the angle formed by $l_2$ and the $x$ axis, and call $\rho$ the angle formed by $l_2$ and the $y$ axis.
Then $\theta = \beta$ and $l_1 = l_2$.
Now reflect $b$ about the $y$ axis and call that point $c$.
Draw $l_3$ from the origin to $c$ and call $\phi$ the angle formed by $l_3$ and the $y$ axis.
Then $l_2 = l_3$ and therefore $l_3 = l_1$.
Since the $x$ and $y$ axes are orthogonal, $\theta$ and $\beta$ are the complements of $\phi$ and $\rho$, therefore $\theta + \beta + \phi + \rho = 180$ degrees, and $l_1 + l_3$ is the diameter of the circle with the origin as center.
Therefore $c$ is symmetric to $a$ with respect to the origin.
I strongly suggest drawing this out.
