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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