Draw An angle of 135

Given: A ray OA.
Required: To construct an angle of 135⁰ at O.
Steps of construction :

1. Produce AO to A' to form OA'.
2. Taking O as centre and some radius, draw an arc of a circle, which intersects OA at a point B and OA' at a point B'.
3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.
4. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
   
5. Draw the ray OE passing through C.
   Then $\angle$EOA = 60^o.
6. Draw the ray OF passing through D.
   Then $\angle$FOE = 60^o.
7. Taking C and D as centres and with the radius more than $1 \over2$CD, draw arcs to intersect each other, say at G.
8. Draw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE, i.e. $\angle$FOG = $\angle$EOG = $1 \over2$$\angle$FOE = $1 \over2$(60^o) = 30^o.
   Thus $\angle$GOA = $\angle$GOE + $\angle$EOA = 30^o + 60^o = 90^o
   $\angle$B'OH = 90^o
9. Taking B' and H as centres and with the radius more than $1 \over2$B'H, draw arcs to intersect each other, say at I.
10. Draw the ray OI. This ray OI is the bisector of the angle B'OG i.e. $\angle$B'OI = $\angle$GOI = $1 \over2$$\angle$B'OG = $1 \over2$(90^o) = 45^o
    $\angle$IOA = $\angle$IOG + $\angle$GOA
    = 45^o + 90^o = 135^o
    On measuring the $\angle$IOA by protractor, find that $\angle$IOA = 135^o.