# Construct a

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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