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

To do:

We have to construct an angle of $45^o$ at the initial point of a given ray and justify the construction.

Solution:

Steps of construction:

(a) Draw a ray $BC$.

(b) With center $B$ and a suitable radius draw an arc meeting $BC$ at $E$.

(c) With center $E$ and the same radius as above draw an arc intersecting the above arc at $F$

(d) With center $F$ and the same radius as above draw an arc intersecting the above arc at $G$

(e) With $E$ and $F$ as centers and radius more than $\frac{1}{2}EF$, draw two arcs meeting each other at $H$.

(f) Join $BH$ and produce it to $X$ so that $\angle XBC = 90^o$.

(g) Let $BH$ intersects the first arc at $D$

(h) With $D$ and $E$ as centers and radius greater than $\frac{1}{2}DE$, draw two arcs meeting each other at $K$

(i) Join $BK$ and extend it to $A$ forming ray $BA$

(j) Therefore, $\angle CBA=45^o$

Justification:

We have,

$\angle HBC=90^o$

We also have a perpendicular bisector $BA$ which divides $\angle HBC$

This implies,

$ABC=\frac{1}{2}HBC$

$ABC=\frac{1}{2}(90^o)$

$ABC=45^o$

Therefore, justified.

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Updated on: 10-Oct-2022

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