If the areas of two similar triangles are equal, prove that they are congruent.

Let's consider two similar triangles ΔABC and ΔDEF

Thus, ΔABC ∼ ΔDEF

⇒ AB/DE = BC/EF = CA/FD (SSS criterion)

But, Area of ΔABC = Area of ΔDEF (According to the given question)

⇒ Area of ΔABC / Area of ΔDEF = 1 ................. (1)

But,

Area of ΔABC / Area of ΔDEF = (AB)² / (DE)² = (BC)² / (EF)² = (CA)² / (FD)² (According to theorem 6.6) ............... (2)

From equation (1) and (2),

(AB)² / (DE)² = 1

⇒ (AB)² = (DE)²

⇒ AB = DE ............... (3)

Similarly,

⇒ BC = EF ....(4)

⇒ CA = FD .....(5)

Now, 

In ΔABC and ΔDEF

⇒ AB = DE [from equation(3)]

⇒ BC = EF [from equation (4)]

⇒ CA = FD [from equation (5)]

Thus, ΔABC ≅ ΔDEF (SSS congruency)