**Menelaw’s Theorem**

**Theorem: **If a transversal cut the sides BC, CA, AB of a triangle in D, E, F
respectively then

BD/DC. CE/EA. AF/FB = -1.

**Proof:**

Let A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) be

Let the transversal be ax + by + c = 0

BD/DC = The ratio in which ax + by + c = 0 divides BC

= – (ax₂ + by₂ + c)/ (ax₃ + by₃ + c)

CE/EA = The ratio in which ax + by + c = 0 divides CA

= – (ax₃ + by₃ + c)/ (ax₁ + by₁ + c)

AF/FB = The ratio in which ax + by + c = 0 divides AB

= – (ax₁ + by₁ + c)/ (ax₂ + by₂ + c)

Therefore, BD/DC . CE/EA . AF/FB

= – (ax₂ + by₂ + c)/ (ax₃ + by₃ + c) . – (ax₃ + by₃ + c)/ (ax₁ + by₁ + c) . – (ax₁ + by₁ + c)/ (ax₂ + by₂ + c) = -1