Question

Construct an angle of $ {45^ \circ } $ at the initial point of a given ray and justify the construction.

Answer 1

Hint: We will construct the angle $ {45^ \circ } $ by drawing a ray initially. Then we will construct number of rays to get our angle. We will justify our result with help of congruence concepts. We will use SSS congruence to justify the result.

Complete step-by-step answer:
Let us consider the given ray as $ OA $ .
Step 1
 We need to construct an angle of $ {45^ \circ } $ at $ O $ and then we will justify the construction.
Step 2
We will take $ O $ as centre and consider a particular radius to draw an arc, which will intersect $ OA $ at point $ B $ .
Step 3
Now we will take $ B $ as a centre and with radius considered above we will again draw an arc which will intersect the arc that we have already drawn at a point $ C $ .
Step 4
Now, we will take $ C $ as a centre and with radius considered above we will again draw an arc which will intersect the arc that we have already drawn at a point $ D $ .
Step 5
We draw another ray $ OE $ passing through the point $ C $ . From the construction the angle $ \angle EOA = {60^ \circ } $ .
Step 6
Now we draw a ray $ OF $ passing through the point $ D $ . From the construction the angle $ \angle FOE = {60^ \circ } $ . Next, we will take $ C $ and $ D $ as centres and will consider radius equals to more than $ \dfrac{1}{2}CD $ , we will draw arcs which will intersect as point $ G $ .
Then we will draw a ray $ OG $ such that $ OG $ is the bisector of the angle $ \angle FOE $ . This can be expressed as:
 $ \begin{array}{c}
\angle EOG = \dfrac{1}{2}\angle FOE\\
 = \dfrac{1}{2}\left( {{{60}^ \circ }} \right)\\
 = {30^ \circ }
\end{array} $
Hence,
 $ \begin{array}{c}
\angle GOA = \angle GOE + \angle EOA\\
 = {30^ \circ } + {60^ \circ }\\
 = {90^ \circ }
\end{array} $
Then we will take $ O $ as centre and considering any radius, we will draw an arc which will intersect the rays $ OA $ and $ OG $ at a point $ H $ and point $ I $ respectively.
Now we will take $ H $ and $ I $ as centres and will consider radius equals to more than $ \dfrac{1}{2}HI $ , we will draw arcs that intersect each other at another point $ J $ .
Next, we will draw the ray $ OJ $ such that $ OJ $ is the bisector of the angle $ \angle GOA $ . This can be expressed as:
\[\begin{array}{c}
\angle GOA = \angle AOJ\\
 = \dfrac{1}{2}\angle GOA\\
 = \dfrac{1}{2}\left( {{{90}^ \circ }} \right)\\
 = {45^ \circ }
\end{array}\]
Hence, we have constructed the required angle.

Now for the justification we join $ BC $ , from the construction we know that $ OB = OC = BC $ .
Since all the sides of the triangle $ \Delta COB $ are equal. Hence it is an equilateral triangle. From the properties of equilateral triangles, we know that all the angles of equilateral triangles are $ {60^ \circ } $ .
Therefore $ \angle COB = {60^ \circ } $ and $ \angle EOA = {60^ \circ } $ .
Now we will join $ CD $ , from the construction we know that $ OD = OC = CD $ .
Since, all the sides of the triangle $ \Delta DOC $ are equal. Hence it is an equilateral triangle. From the properties of equilateral triangles, we know that all the angles of equilateral triangles are $ {60^ \circ } $ .
Therefore, $ \angle DOC = {60^ \circ } $ and $ \angle FOE = {60^ \circ } $ . Then we will join $ CG $ and $ DG $ .
Now, in $ \Delta ODG $ and $ \Delta OCG $ ,
The $ OD $ and $ OC $ are the radius of the same arc. Hence
 $ OD = OC $
Also $ DG $ and $ CG $ are arcs of equal radii. Hence,
 $ DG = CG $
The common side of the triangles is $ OG $ .
Since all the corresponding sides are equal in $ \Delta ODG $ and $ \Delta OCG $ . We can say that from SSS congruence that is $ \Delta ODG \cong \Delta OCG $ .
Therefore $ \angle DOG = \angle COG $ as both the angles are the concurrent part of the congruent triangles. This can also be expressed as:
 $ \begin{array}{c}
\angle FOG = \angle EOG\\
 = \dfrac{1}{2}\left( {{{60}^ \circ }} \right)\\
 = {30^ \circ }
\end{array} $
Hence, $ \begin{array}{c}
\angle GOA = \angle GOE + \angle EOA\\
 = {30^ \circ } + {60^ \circ }\\
 = {90^ \circ }
\end{array} $
Now we will join $ HJ $ and $ IJ $ , also consider $ \Delta OIJ $ and $ \Delta OHJ $ ,
The $ OI $ and $ OH $ are the radius of the same arc. Hence,
 $ OI = OH $
Also $ IJ $ and $ HJ $ are arcs of equal radii. Hence,
 $ IJ = HJ $
The common side of the triangles is $ 0J $ .
Since all the corresponding sides are equal in $ \Delta OIJ $ and $ \Delta OHJ $ . We can say that from SSS congruence that is $ \Delta OIJ \cong \Delta OHJ $ .
Therefore $ \angle IOJ = \angle HOJ $ as both the angles are the concurrent part of the congruent triangles.
This can also be expressed as:
 $ \begin{array}{c}
\angle AOJ = \angle GOJ\\
 = \dfrac{1}{2}\angle GOA\\
 = \dfrac{1}{2}\left( {{{90}^ \circ }} \right)\\
 = {45^ \circ }
\end{array} $
Hence our result is justified.

Note: We will have to be very cautious while constructing the rays and angles, as a minor mistake can misinterpret the result. This question is based on the precise use of compass along with the assumption that we have to make during the construction.