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°.
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)
∠ FID = ∠ FAC = 90°
[from (5) and (6)]
Hence, the constructed angle at A is 90°.
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