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

Steps of construction:

Step 1: A ray OP is drawn.
Taking O as a centre and any radius draw an arc cutting OP at Q.

Step 2: Taking Q as a centre and any radius draw an arc cutting previous arc at R.
Repeat the process with R to cut the previous arc at S.

Step3: Taking R and S as centre draw the arc of radius more tha half of RS
and draw two arcs intersecting at A.
Join OA

Hence ∠POA = 90^°
Justification:

We need to prove ∠POA = 90^°
Join OR and OS and RQ.

By construction
OQ = OS = QR
So, ΔROQ is an equilateral triangle.
Similarly ΔSOR is an equilateral triangle.
So,
∠SOR = 60^°
As ∠ROQ = 60^°
It means ∠ROP = 60^°
Join AS and AR.

Now in ΔOSA and ΔORA,
SR = SR (common)
AS = AR (radii of same arcs)
OS = OR (radii of same arcs)
ΔOSA ≅ ΔORA
∠SOA = ∠ROA (CPCT)
So,
∠SOA = ∠ROA = 1/2 ∠SOR
= 1/2 (60^°)
= 30^°
Now,
∠POA = ∠ROA + ∠POR
= 30^° + 60^°
= 90^°

hence Justified