(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 \(\frac{DE}{2}\) .
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 \(\frac{DG}{2}\) .
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.
∠EAF = ∠DAF = \(\frac{1}{2}\)∠DAE = 90°.
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.
∠GAH = ∠DAH = \(\frac{1}{2}\)∠DAG = \(\frac{1}{2}\)∠DAF = 45°
Hence,
∠DAI =∠DAH = 45°
(ii) 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 \(\frac{DE}{2}\) .
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.
∠EAF = ∠DAF = \(\frac{1}{2}\)∠DAE = 90°
Hence,
∠DAG =∠DAF = 90°.
