A bit of physical reasoning should convince you that this must be true. First, recall two familiar properties of
water: (i) it has virtually no resistance to lateral (**shear**) forces at its surface (for example,
however gently you blow across the surface of water, it will move), and (ii) it is very nearly an incompressible
fluid --- it is very difficult to force water molecules closer together. Now consider a thin square horizontal
slab of water with vertical sides which is embedded within a larger body of water, as shown in Figure 1.

Assume static equilibrium, i.e. no part of the water is in motion relative to any other part. Lack of resistance to shear
(property (i)) implies that the external force on each slab face must act in a direction **perpendicular**
to that face. If this were not so, there would be a shear stress across the slab face and it would move, which
would contradict the assumption of static equilibrium. The forces on the slab faces are therefore as shown in
Figure 1. Not shown are the two other forces which act on the slab edges that are in the plane of the figure.

Continuing with the requirement for equilibrium, it is clear that opposing horizontal forces, such as F1 and F2 must be equal in magnitude. Otherwise the slab would move sideways. Moreover, there is nothing special about the directions of F1 and F2. By symmetry, if you rotate the slab to an arbitrary new position in the horizontal plane, the new lateral forces on the slab edges must be the same as before. This also follows from the incompressibility of water (property (ii)): squeezing the slab in one direction will make it bulge in the other direction until all forces are balanced.

Conclusion: the horizontal forces (and pressures) acting at a point from any direction **must be identical**
in a static fluid.

What about vertically? If the gravitational acceleration were not shown in Figure 1, you could imagine that you were
looking *down* at the slab, and then use exactly the same arguments to prove that F3=F4=F1=F2 (the reasoning
did not depend on slab shape!). So let's remove the complicating effect of the gravitational acceleration: let the
thickness of the slab approach zero! This forces the mass of the slab, and therefore the gravitational force on it,
to both approach zero, and allows us to exploit the symmetry of the situation. Finally, letting the length and width
of the slab also approach zero gives us: **the pressure at a point in a static body of water is exactly the
same from all directions**.

There is also a simple, yet elegant, mathematical proof of the above. Consider the water-filled vessel shown in Figure 2, and imagine a prism-shaped element of water (as shown).

The prism's slanted side has length S, is inclined at the angle A, and the prism has length L in the direction perpendicular to the plane of the figure. As before, static equilibrium requires that the forces on the prism faces be perpendicular to the faces. Using the notation in Figure 2, translational equilibrium in the horizontal direction requires that:

F2 = F1*sinAEquation 1

But the forces F1 and F2 act on surface areas A1 and A2, respectively, where

A1=L*S ; A2=L(S*sinA)Equation 2

F1 = P1*A1 = P1*L*S ; F2 = P2*A2 = P2*L(S*sinA)Equation 3

where P1 and P2 are the pressures corresponding to F1 and F2, respectively. Substituting from Equation 3 into Equation 1, we get

P1 = P2Equation 4

**regardless of the direction of the force F1.**
For translational equilibrium vertically we must have

F1*cosA + Mg = F3Equation 5

where M is the mass of the water prism and g is the acceleration due to gravity. The mass M is simply the density (d) of water multiplied by the volume of the prism:

M = d[0.5L(S*sinA)(S*cosA)]Equation 6

But the force F3 acts on a surface of area

A3 = L(S*cosA)Equation 7

so the upward pressure is P3=F3/A3, or

F3 = P3*L(S*cosA)Equation 8

Substituting from Equations 3 and 8 into Equation 5 gives

P3 = P1 + 0.5*d(S*sinA)gEquation 9

Since we imposed no constraint on the length S of the prism's sloping side, or the angle A it makes with the horizontal, we can now let S approach zero, shrinking the prism to a point and giving

P3 = P1Equation 10

at that point, **regardless of the direction of the force F1**.

Equations 4 and 10 together state that the horizontal and vertical pressures exerted at a point are equal to the pressure exerted at that point from any other direction. The proof is complete.