Introduction:
The statement "For angles of projection which exceed or fall short of 45degree by an equal amount the ranges are equal" implies that if the angle of projection of a projectile is either greater than or less than 45 degrees by the same amount, then the range of the projectile will remain the same. In other words, the range of a projectile is independent of the angle of projection if the angles differ from 45 degrees by the same amount.

Proof:
To prove this statement, we can use the following formula for the range of a projectile:
Range = (v^2/g) * sin(2θ)

Where v is the initial velocity of the projectile, g is the acceleration due to gravity, and θ is the angle of projection.

Case 1: Exceeding 45 degrees by an equal amount:
Let's assume that the angle of projection is 45 + x degrees, where x is any positive angle. Therefore, the angle of projection can be expressed as:

θ1 = 45 + x

Using the above formula for range, we get:

Range1 = (v^2/g) * sin(2θ1)

Range1 = (v^2/g) * sin(2(45 + x))

Range1 = (v^2/g) * [sin(90 + 2x)]

Range1 = (v^2/g) * cos(2x)

Case 2: Falling short of 45 degrees by an equal amount:
Let's assume that the angle of projection is 45 - x degrees, where x is any positive angle. Therefore, the angle of projection can be expressed as:

θ2 = 45 - x

Using the above formula for range, we get:

Range2 = (v^2/g) * sin(2θ2)

Range2 = (v^2/g) * sin(2(45 - x))

Range2 = (v^2/g) * [sin(90 - 2x)]

Range2 = (v^2/g) * cos(2x)

Conclusion:
From the above calculations, we can see that the ranges of the projectile in both cases are equal and are given by the formula:

Range = (v^2/g) * cos(2x)

Therefore, we can conclude that for angles of projection which exceed or fall short of 45 degrees by an equal amount, the ranges are equal.