Question

Prove Theorem $9.10$ That is, show that if a graph in the plane has any two of three types of symmetry, namely, symmetry about the $x-$axis, symmetry about the $y-$axis, and symmetry about the origin, then it has the third type of symmetry as well.

Short answer

Proof: $c$is symmetric to $a$with respect to the origin.

Step-by-step solution

Given information

It has the third type of symmetry as well.

Calculation

Take a point $a$ in the first quadrant.

Draw a line $l_1$from the orlgin to $a$and let the angle formed by $l_1$and the $x$axis.

Now reflect the point $a$ about $x$ axis and call it as point $b$

Draw a 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 $l_1=l_2$

Calculation

Now reflect$b$ about $y$ axis and call that paint $c$

Draw $l_3$from the origin to $c$and call $\varphi$the angle formed by $l_3$and $y-$axis.

Then $I_2=I_3$and therefore $I_3=l_1$

Since $x,y$axes are orthogonal, $\theta$and $\beta$are complements of $\varphi$and $\rho$

Therefore, $\theta +\beta +\varphi +\beta =180,l_1+l_3$the diameter of the circle with the origin as the center. Therefore, $c$is symmetric to $a$with respect to origin.

Hence the proof.