Question

construct an angle of 45^0 at the initial point of a given ray
and justify the construction ​

Answer 1

$\sf \large \blue{Construction}$

1. Draw a ray OA

2. Taking O as center and any radius, draw an arc cutting OA at B

3. Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C

4. With C as center and the same radius, draw an arc cutting the arc at D

5. With C and D as centers and radius more than ½ CD, draw two arcs intersecting at P

6. Join OP

Thus, ∠AOP = 90°

Now, we draw bisector of ∠AOP

7. Let OP intersect the original arc at point Q

8. Now, taking B and Q as centers and radius greater than ½ BQ, draw two arcs intersecting at R

9. Join OR

Thus, ∠AOR = 45°

_________________◆____________________

$\sf \large \blue{Justification}$

We need to prove ∠AOR = 45°

Proof:

Join OC and OB

Thus, OB = BC = OC (Radius of equal arcs)

∴ ∆OCD is an equilateral triangle

∴ ∠BOC = 60°

Join OD and CD

Thus, OD = OC = CD (Radius of equal arcs)

∴ ∆DOC is an equilateral triangle

∴ ∠DOC = 60°

Join PD and PC

Now, in ∆ODP and ∆OCP,

OD = OC (Radius of same arcs)

DP = CP (Arc of same radii)

OP = OP (Common)

∴ ∆ODP ≅ ∆OCP (By SSS congruency)

∴ ∠DOP = ∠COP (corresponding parts of congruent triangle)

So, we can say that

∠DOP = ∠COP = ½ ∠DOC

∠DOP = ∠COP = ½ × 60° = 30°

Now, ∠AOP = ∠BOC + ∠COP

∠AOP = 60° + 30° = 90°

Join QR and BR

In ∆OQR and ∆OBR,

OQ = OB (Radius of same arcs)

QR = BR (Arc of same radii)

QR = OR (Common)

∴ ∆OQR ≅ ∆OBR (By SSS congruency)

∴ ∠QOR = ∠BOR (corresponding parts of congruent triangle)

∠QOR = ∠BOR = ½ ∠AOP

∠DOP = ∠COP = ½ × 90° = 45°

Thus, ∠AOR = 45°

Hence, justified