Construct the following angles at the initial point of a given ray and justify the construction :

(i) 45° (ii) 90°

i) The steps of the required construction are:

1) Draw a line segment BC. Taking any arbitrary point, A on line segment BC as the the center and any radius draw a semicircle, intersecting BC at points D and E.

2) Taking D as the center draw an arc of any radius greater than . Now, Taking E as the center and the keeping the same radius, draw another arc, intersecting the previous arc at F. Join AF, which intersects the semicircle at point G.

3) Taking D as the center draw an arc of any radius greater than . Now, Taking G as the center and the keeping the same radius, draw another arc, intersecting the previous arc at H. Join AH, which intersects the semicircle at point I.

4) ∠DAI=45°.

Justification:

Since DAE is a straight line therefore ∠DAE=180°.

Consider ∆EAF and ∆DAF

AE=AD (Radius of semi-circle)

EF=DF (By construction)

AF=AF (Common side)

Hence, By SSS criteria, ∆EAF≅∆DAF.

Therefore, by C.P.C.T. .

Consider ∆GAH and ∆DAH

AG=AD (Radius of semi-circle)

GH=DH (By construction)

AH=AH (Common side)

Hence, By SSS criteria, ∆GAH≅∆DAH.

Therefore, by C.P.C.T.

Hence, ∠DAI=∠DAH=45°.

ii) The steps of the required construction are:

The steps of the required construction are:

1) Draw a line segment BC. Taking any arbitrary point, A on line segment BC as the the center and any radius draw a semicircle, intersecting BC at points D and E.

2) Taking D as the center draw an arc of any radius greater than . Now, Taking E as the center and the keeping the same radius, draw another arc, intersecting the previous arc at F. Join AF, which intersects the semicircle at point G.

3) ∠DAG=90°.

Justification:

Since DAE is a straight line therefore ∠DAE=180°.

Consider ∆EAF and ∆DAF

AE=AD (Radius of semi-circle)

EF=DF (By construction)

AF=AF (Common side)

Hence, By SSS criteria, ∆EAF≅∆DAF.

Therefore, by C.P.C.T.

Hence, ∠DAG=∠DAF=90°.