Question

Construct an angle of $45°$ at the initial point of a given ray and justify the construction.

Answer 1

Step 1: Constructing an angle of $90°$

Let us draw a ray $\mathrm{OA}$ and considering $\mathrm{O}$ as a centre with any radius, draw an arc that cuts $\mathrm{OA}$ at $\mathrm{B}$.

Considering $\mathrm{B}$ as a centre with the same radius as before, mark a point $\mathrm{C}$ and then considering $\mathrm{C}$ as a centre and the same radius, mark a point $\mathrm{D}$ on the previously drawn arc.

Now let us take $\mathrm{C}$ and $\mathrm{D}$ as the centre and radius more than $\frac{1}{2}\mathrm{CD}$, draw two arcs that intersect each other at $\mathrm{P}$.

Now the ray $\mathrm{OP}$ is joined which makes an angle of $90°$.

Thus, $\angle \mathrm{AOP}=90°$ .

Step 2: Bisecting the angle of $90°$

Let $\mathrm{OP}$ intersects the original arc at $\mathrm{Q}$.

Considering $\mathrm{B}$ and $\mathrm{Q}$ as centre and radius greater than $\frac{1}{2}\mathrm{BQ}$, draw two arcs intersecting at point $\mathrm{R}$. Join $O\mathrm{R}$.

Thus, $\angle \mathrm{AOR}=45°$.

Justification

From the construction,

$\angle \mathrm{AOP}=90°$

The perpendicular bisector from the point $\mathrm{B}$ and $\mathrm{Q}$, divides the $\angle \mathrm{AOP}$ into two halves. So it becomes

$\angle \mathrm{AOR}=$ $\frac{1}{2}\angle \mathrm{AOP}$

$\angle \mathrm{AOR}=$$\frac{1}{2}\times 90°=45°$

Hence Proved.