In the given figure, AD = DB = DE, ∠EAC = ∠FAC and ∠F = 90°

Prove that :

(i) ∠AEB = 90°

(ii) △ CEG is isosceles

(iii) ∠CEG = ∠EAF.

(i) Given: AD = DB = DE

∠EAC = ∠FAC

∠F = 90°

Since BD = DE, the angles opposite to equal sides are also equal:

∠DBE = ∠CEA = x

Similarly, since DE = DA, we have:

∠DEA = ∠DAE = y

In Δ AEB, using the angle sum property:

⇒ ∠AEB + ∠ABE + ∠EAB = 180°

⇒ (x + y) + x + y = 180°

⇒ 2(x + y) = 180°

⇒ x + y = $\dfrac{180°}{2}$

⇒ x + y = 90°

⇒ ∠AEB = 90°

Hence, ∠AEB = 90°.

(ii) In Δ AEF, using the angle sum property:

⇒ ∠AEF + ∠AFE + ∠EAF = 180°

⇒ ∠AEF + 90° + ∠EAF = 180°

⇒ ∠AEF + ∠EAF = 180° - 90°

⇒ ∠AEF + ∠EAF = 90°

⇒ ∠AEF = 90° - ∠EAF ……………….(1)

Since, ∠AEB = ∠AEC = 90°,

⇒ ∠AEF + ∠FEC = 90°

Substituting Equation (1):

⇒ 90° - ∠EAF + ∠FEC = 90°

⇒ ∠EAF = ∠FEC ………………… (2)

From the figure,

∠FEC = ∠CEG

Using equation (2), we get:

⇒ ∠CEG = ∠EAF ………………..(3)

Since △ CEG has two equal angles, it is isosceles.

Hence, △ CEG is isosceles.

(iii) From equation (3), ∠CEG = ∠EAF.

Hence, ∠CEG = ∠EAF.