Construct an angle of \(45^\circ\) at the initial point of a given ray and justify the construction.

Steps of construction:



1. Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then, \(\angle{EOA}\) = \(60^\circ\)

5. Draw the ray OF passing through D. Then, \(\angle{FOE}\) = \(60^\circ\).

6. Next, taking C and D as centres and with the radius more then (\(\frac{1}{2} \) )CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the \(\angle{FOE}\)

\(\Rightarrow \) \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (\(\frac{1}{2} \) )\(60^\circ\) = \(30^\circ\)
\(\therefore \) \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)
\(\Rightarrow \) \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

8. Now taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I, respectively.

9. Next, taking H and I as centres and with the radius more than (\(\frac{1}{2} \) )HI, draw arcs to intersect each other, say at J.

10. Draw the raw OJ. This ray OJ is the required bisector of the \(\angle{GOA}\).

Thus, \(\angle{GOJ}\) = \(\angle{AOJ}\) = (1/2)\(\angle{GOA}\) = (\(\frac{1}{2} \) )\(90^\circ\) = \(45^\circ\)


Justification:
i)Join BC.
Then, OC = OB = BC ...(By construction)
\(\therefore \) , \(\angle{COB}\) is an equilateral triangle.
Thus, \(\angle{COB}\) = \(60^\circ\)
Also, \(\angle{EOA}\) = \(60^\circ\)

ii)Join CD.
Then, OD = OC = CD ...(By construction)
Therefore, \(\angle{DOC}\) is an equilateral triangle.
Thus, \(\angle{DOC}\) = \(60^\circ\)
Also, \(\angle{FOE}\) = \(60^\circ\)

iii)Now, join CG and DG.
In \(\triangle{ODG}\) and \(\triangle{OCG}\), we have,
OD = OC ...((Radii of the same arc)
DC = CG ...(Arcs of equal radii)
and OG = OG ...(Given)
By SSS criterion, we get,
\(\triangle{ODG}\) \(\displaystyle \cong\) \(\triangle{OCG}\)
Hence, \(\angle{DOG}\) = \(\angle{COG}\) ...(By CPCT)
So, \(\angle{FOG}\) = \(\angle{EOG}\) = (\(\frac{1}{2} \) )\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)
Thus, \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)
i.e., \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

iv) Now, join HJ and IJ.
In \(\triangle{OIJ}\) and \(\triangle{OHJ}\), we have,
OI = OH ...((Radii of the same arc)
IJ = HJ ...(Arcs of equal radii)
and OJ = OJ ...(Given)
By SSS criterion, we get,
\(\triangle{OIJ}\) \(\displaystyle \cong\) \(\triangle{OHJ}\)
Hence, \(\angle{IOJ}\) = \(\angle{HOJ}\) ...(By CPCT)
So, \(\angle{AOJ}\) = \(\angle{GOJ}\) = (1/2)\(\angle{GOA}\) = (\(\frac{1}{2} \) )\(90^\circ\) = \(45^\circ\).