To construct an angle of 90°, the given steps can be followed
1. Draw a ray OA
2. Considering O as the centre with any radius, we draw an arc DCB is that cuts OA at B.
3. Considering B as a centre with the same radius, mark a point C on the arc DCB.
4. Considering C as a centre and the same radius, mark a point D on the arc DCB.
5. Let us take C and D as the centre, draw two arcs that intersect each other with the same radius at P.
6. Now join the ray OP making an angle of 90° with OP.
To prove ∠POA = 90°
To prove this, draw a dotted line from the point O to C and O to D and the angles formed are:
From the construction, we note that
OB = BC = OC
Hence, OBC is an equilateral triangle
So that, ∠BOC = 60°.
OD = DC = OC
Hence, DOC is an equilateral triangle
So that, ∠DOC = 60°.
From SSS triangle congruence rule
△OBC ≅ OCD
So, ∠BOC = ∠DOC [By C.P.C.T]
Hence, ∠COP = ½ ∠DOC = ½ (60°).
∠COP = 30°
To determine ∠POA = 90°:
∠POA = ∠BOC+∠COP
∠POA = 60°+30°
∠POA = 90°