Does having two lines of symmetry $y=0$ and $y=-x$ imply that the shape also has lines of symmetry $x=0$ and $y=x$
Let $(u,v)$ be a point on the curve. Then:
- symmetry about $y=0$ implies that $(u,-v)$ is on the curve;
- symmetry about $y=-x$ implies that $(-v,-u)$ is on the curve.
Now we can apply these rules iteratively:
- reflect $(u,-v)$ about $y=-x$ to get $(v,-u)$
- reflect $(v,-u)$ about $y=0$ to get $(v,u)$
And $(v,u)$ is $(u,v)$ reflected about $y=x$. So $y=x$ is indeed a line of symmetry.
Similarly, reflecting $(-v,-u)$ about $y=0$ then $y=-x$ gives $(-u,v)$, which is $(u,v)$ reflected about $x=0$. So $x=0$ is also a line of symmetry.
Yes, this is true.
Consider a point $(x,y)$ in the $1^{\text{st}}$ quadrant. By line of symmetry $y=0$, the curve also contains $(-x,y)$. Then by the line of symmetry $y=-x$, the curve contains $(-x,-y)$ (since it contains $(x,y)$). Now, the curve must contain $(-x,y)$ since the line of symmetry $y=0.$
Therefore we have proved the $4$ points $(x,y),(x,-y),(-x,-y),(-x,y)$ are simultaneously on the curve or off the curve.
Then you can try prove it has lines of symmetry $x=0$ and $y=x$. $(*)$
Comment if you got stuck on how to prove $(*)$ or if I have made anything unclear.