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 {tex}\angle{/tex}EOA = 60^o.
6. Draw the ray OF passing through D.
   Then {tex}\angle{/tex}FOE = 60^o.
7. Taking C and D as centres and with the radius more than {tex}1 \over2{/tex}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. {tex}\angle{/tex}FOG = {tex}\angle{/tex}EOG = {tex}1 \over2{/tex}{tex}\angle{/tex}FOE = {tex}1 \over2{/tex}(60^o) = 30^o.
   Thus {tex}\angle{/tex}GOA = {tex}\angle{/tex}GOE + {tex}\angle{/tex}EOA = 30^o + 60^o = 90^o
   {tex}\angle{/tex}B'OH = 90^o
9. Taking B' and H as centres and with the radius more than {tex}1 \over2{/tex}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. {tex}\angle{/tex}B'OI = {tex}\angle{/tex}GOI = {tex}1 \over2{/tex}{tex}\angle{/tex}B'OG = {tex}1 \over2{/tex}(90^o) = 45^o
    {tex}\angle{/tex}IOA = {tex}\angle{/tex}IOG + {tex}\angle{/tex}GOA
    = 45^o + 90^o = 135^o
    On measuring the {tex}\angle{/tex}IOA by protractor, find that {tex}\angle{/tex}IOA = 135^o.