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# 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)$$\angle{GOA}$$ = ($$\frac{1}{2}$$ )$$90^\circ$$ = $$45^\circ$$.