If lines O A_{0}, O A_{1}, O A_{2}, etc., be drawn from any fixed point 0, then the areas of the triangles O A_{0} A_{1}, O A_{1} A_{2}, O A_{2} A_{3}, 0 A_{3} A_{4}, will be all equal. For each area is one-half the product of the base of the triangle into the perpendicular O T from O on A_{0} A_{1}, and, as the bases of all the triangles are equal, it follows that their areas are equal. Thus we learn that a particle moving without the action of force will describe around any fixed point O equal areas in equal times. [Illustration: Fig. 60.—FIRST LAW OF MOTION EXEMPLIFIES CONSTANT MOMENT OF MOMENTUM.] The product of the mass of the particle and its velocity is termed the momentum. If the momentum be multiplied by O T the product is termed the moment of momentum around O. We have in this case the simplest example of the important principle known as the conservation of moment of momentum. The moment of momentum of a system of particles moving in a plane is defined to be the excess of the sum of the moments of momentum of those particles which tend round O in one direction, over the sum of the moments of momentum of those particles which tend round O in the opposite direction. If we deem those moments in one direction round O as positive, and those in the other direction as negative, then we may say that the moment of momentum of a system of particles moving in a plane is the algebraical sum of the several moments of momentum of each of the particles. § 9. A GEOMETRICAL PROPOSITION. The following theorem in elementary geometry will be required:— [Illustration: Fig. 61.—A USEFUL GEOMETRICAL PROPOSITION.] Let A B and A C be adjacent sides of a parallelogram, Fig. 61, of which A D is the diagonal, and let O be any point in its plane. Then the area O A C is the difference of the areas O A D and O A B. Draw D Q and C P parallel to O A. Then O A D = O A Q, whence O A D – O A B = O B Q = O A P = O A C. § 10. RELATION BETWEEN THE CHANGE OF MOMENT OF MOMENTUM AND THE FORCE ACTING ON THE PARTICLE. [Illustration: Fig. 62.—ACCELERATION OF MOMENT OF MOMENTUM EQUALS MOMENT OF FORCE.] Let A_{1} and A_{2}, Fig. 62, be two adjacent points on the path of the particle, and let A_{1} Q and A_{2} R be the tangents at those points. Let S Q represent the velocity of the particle at A_{1}, and SR the velocity of the particle at A_{2}. Then Q R represents both in magnitude and direction the change in velocity due to the force F, which we suppose constant both in magnitude and direction, while the particle moves from A_{1} to A_{2} in the small time _t_; we have also Q R = F _t_ ÷ _m_.