Newton's Third Law of Motion

Statement

To every action there is always an equal and opposite reaction

Proof
Consider about two moving bodies in an isolated system. Masses of the bodies are M1 and M2. Velocities are V1 and V2.

Due to interaction, their velocities change. Consequently there will be change in their momenta in time ∆t. Suppose, change in momentum of the first body = ∆p1 and change in momentum of the second body = ∆p2.

From the law of conservation of linear momentum, we know that

∆p1 + ∆p2 = 0
∆p2 = -∆p1
Now, dividing both sides by ∆t
∆p2/∆t = - ∆p1/∆t
Take limit both sides when ∆t tends to zero.
dp2/dt = - dp1/dt
F2 = - F1
Force acting on M2 = - Force acting on M1
Action = - Reaction
To every action there is always an equal and opposite reaction. Proved.
This represents Newton's third law of motion.

Thales Theorem

Statement


"If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio "
Given
To Prove that
Construction
Join B to E and C to D. Draw perpendicular EN on AB and DM on AC.
Proof


ABC is a triangle. DE parallel to BC.

AD/DB = AE/EC

Consider in triangle ADE and BDE.
ar(ADE) = 1/2 ADxEN
ar(BDE) = 1/2DBxEN
Therefore,
ar(ADE)/ar(BDE) = AD/DB ......... (i)
Again, consider in triangle ADE and CDE.
ar(ADE) = 1/2 AExDM
ar(CDE) = 1/2 ECxDM
Therefore,
ar(ADE)/ar(CDE) = AE/EC ............ (ii)
Triangles BDE and CDE have same base DE and DE parallel to BC.
Therefore,
ar(BDE) = ar(CDE)
Now from (i) and (ii)
AD/DB = AE/EC Proved.