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3.To construct a regular octagon, construct a square, and locate its centre, O.

Then construct two arcs through O with centres at opposite vertices.

Using the other vertices, draw similar arcs through O.

Join the points located on the square to form an octagon.

Prove the octagon is a regular octagon. We have a square $\displaystyle ABCD$ with center $\displaystyle O$ and side $\displaystyle 2a.$ Code:

A Q P B
* - - o - - - o - - *
| * * |
| * * |
R o * * |
| * * |
| * | 2a
| * O * |
| * * |
| * * |
| * * |
* - - - - - - - - - *
D 2a C

Using center $\displaystyle A$ and radius $\displaystyle r = AO$, locate point $\displaystyle P$.

Using center $\displaystyle B$ and radius $\displaystyle r = BO$, locate point $\displaystyle Q$.

Using center $\displaystyle D$ and radius $\displaystyle r = DO$, locate point $\displaystyle R$.

The radii of all the arcs are equal to: .$\displaystyle r \:=\:AO \:=\:a\sqrt{2}$

. . Hence: .$\displaystyle AP \:=\:BQ \:=\:DR \:=\:a\sqrt{2}$

. . And: .$\displaystyle AQ \:= \:BP \:= \:AR \:= \:(2-\sqrt{2})a$

Then: . $\displaystyle PQ \:=\:2a(\sqrt{2}-1)$

. . This is the length of the four vertical and horizontal sides of the octagon.

We have isosceles right triangle $\displaystyle QAR$ with legs $\displaystyle (2-\sqrt{2})a$

Its hypotenuse is given by: .$\displaystyle QR^2\;=\;[(2-\sqrt{2})a]^2 + [(2-\sqrt{2})a]^2$

. . Then: .$\displaystyle QR^2\;= \;4a^2(3 - 2\sqrt{2}) \;=\;4a^2(\sqrt{2}-1)^2$

So: .$\displaystyle PQ \;=\;2a(\sqrt{2}-1)$

. . This is the length of the four diagonal sides of the octagon.

Hence, the octagon has eight equal sides.

Since each of the four corners has the same isosceles right triangle,

. . all the vertices have 135° angles.

Hence, the octagon has eight equal angles.

Therefore, we have a *regular* octagon.