Show that area I + area Il area Ill. ?

Refer to figure 1 below.

The diagonal of the square bisects one of the interior angle.
θ₁ = θ₂ = 45° - 30° = 15°

Let a be the length of each side of the equilateral triangle.
Let b be the length of each arm of the right triangle III.

In right triangle III:
a² = b² + b²  (Pythagorean theorem)
b² = a²/2

Area III
= (1/2) b²
= (1/2) a²/2
= a²/4 …… [1]

Refer to Figure below.
Move triangle I to the top of triangle II as shown. The two triangles combine to form a isosceles triangle which is shaded.

Area I + Area II
= Area of the shaded triangle
= (1/2) a² sin(θ₁ + θ₂)
= (1/2) a² sin(15° + 15°)
= (1/2) a² (1/2)
= a²/4 …… [2]

[2] = [1]:
Hence, Area I + Area II = Area III