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**
Ex 11.1, 2
**

Construct an angle of 45° 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°
- Now, take B and Q as centers, and radius greater than 1/2 BQ, draw two arcs intersecting at R.
- Join OR.

Thus, ∠ AOR = 45°

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Justification
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We need to prove ∠ AOR = 45°

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Join OC
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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° = 90°

Now,

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Join QR and BR
**

In Δ OQR and Δ OBR

OQ = OB

QR = BR

OR = OR

∴ Δ OQR ≅ Δ OBR

∴ ∠ QOR = ∠ BOR

∠ QOR = ∠ BOR = 1/2 ∠ AOP

∠ DOP = ∠ COP = 1/2 × 90° = 45°

Thus, ∠ AOR = 45°

Hence justified

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