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Ex 11.1, 1
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Construct an angle of 90° at the initial point of a given ray and justify the construction .

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Steps of construction
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- Draw a ray OA.
- Taking O as centre and any radius, draw an arc cutting OA at B.
- Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.
- With C as centre and the same radius, draw an arc cutting the arc at D.
- With C and D as centres and radius more than 1/2 CD draw two arcs intersecting at P.
- Join OP.

Thus, ∠AOP = 90°

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Justification
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We need to prove ∠ AOP = 90°

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Join OC and BC
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Thus,

OB = BC = OC

∴ Δ OCB is an equilateral triangle

∴ ∠ BOC = 60°

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Join OD, OC and CD
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Thus, OD = OC = DC

∴ Δ DOC is an equilateral triangle

∴ ∠ DOC = 60°

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Join PD and PC
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Now,

In Δ ODP and Δ OCP

OD = OC

DP = CP

OP = OP

∴ Δ ODP ≅ Δ OCP

∴ ∠ DOP = ∠ COP

So, we can say that

∠ DOP = ∠ COP = 1/2 ∠ DOC

∠ DOP = ∠ COP = 1/2 × 60° = 30°

Now,

∠ AOP = ∠ BOC + ∠ COP

∠ AOP = 60° + 30°

∠ AOP = 90°

Hence justified