Construct an angle of 90⁰ at the initial point of a given ray and justify the construction.

Given, to construct an angle of 90° from the initial point of a ray and then we should justify if the constructed angle is 90°.

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°.

Justification:

Firstly, join AE, AD, DE, DC, EF and FD.

From the construction, we can say that

AD = DC = AC

Thus, as all the sides are equal ΔADC is an equilateral triangle.

[If all the sides and angles in a triangle are equal then the triangle is an equilateral triangle]

Now, in an equilateral triangle each angle is equal to 60°.

Hence, ∠ DAC = 60° ----- (1)

Similarly, if we consider ΔADF,

AD = DE = EA (by construction)

So, ΔADF is equilateral and ∠ DAE = 60° ----- (2)

From (1) and (2) we can say that, ∠ DAE and ∠ DAC are alternative angles for the line AD, thus making ED parallel to AC.

Now, ED || AC ------- (3)

Also, we know that DF = EF ----- (4)

(As F is formed by the arcs with same radius)

From (4) we can say that, point F lies on the perpendicular bisector of DE.

Thus ∠ FID = 90° ------- (5)

(where I is the point where FA bisects DE.)

Consider the lines AF, AC and DE. Here AF is the transverse to the parallel lines AC and DE.

∠ FID = ∠ FAC ------- (6)

(since the angles are corresponding angles made by the transverse AF with the parallel lines DE and AC)

Thus,

∠ FID = ∠ FAC = 90°

[from (5) and (6)]

Hence, the constructed angle at A is 90°.