Answer :
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
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