Construct an angle of 45⁰ at the initial point of a given ray and justify the construction.

Given, to construct an angle of 45° from the initial point of a ray and then we should justify if the constructed angle is 45°.

Steps of Construction:

i. Draw a ray AB of any length.

ii. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

iii. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

iv. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step ii) at point E.

v. With E and D as centers, and with a radius more than half the length of DE, draw two arcs intersecting at point F.

vi. Join points A and F. The angle formed by FAB is 90°. i.e.

∠ FAB = 90°.

vii. Now, consider G as the point where the first arc (from step ii) meets the line AF and point C where the first arc (from step ii) meets the line AB.

With G and C as centers and with any radius more than half the length of GC, draw two arcs which intersect at point H.

viii. Join points A and H, thus the line forms an angle 45° with the ray AB.

Justification:

Firstly, join GH and CH.

From the construction, we can say that

GH= CH ----- (1)

[as they are arcs of equal radii]

Also,

AG = CG ------ (2)

[as they are the radii of the same arc in step (ii) of the construction ]

For the ΔAGH & ΔACH, side AH is common. ---- (3)

From (1), (2) and (3), we can say that ΔAGH & ΔACH are similar, i.e.

------ (4)

(From SSS property of triangles)

Now, from (4), as the triangles are congruent,

∠ HAG = ∠ HAC ------- (5)

[Since the angles are the corresponding parts of congruent triangles]

But, from step (vi) ∠ FAB = 90°. ----- (6)

Also from the figure, we can clearly say that

∠ HAC + ∠ HAG = ∠ FAB

Now from (5) and (6),

∠ HAC + ∠ HAC = 90°

2 ∠ HAC = 90°

∠ HAC = 45°

Hence, the constructed ∠ HAC is 45°.