Desargues' Theorem

Let $A_1B_1C_1$ and $A_2B_2C_2$ be two triangles. Then points $P=A_1B_1\cap A_2B_2$, $Q=A_1C_1\cap A_2C_2$ and $R=B_1C_1\cap B_2C_2$ are colinear if and only if lines $A_1A_2$, $B_2B_2$, $C_1C_2$ are concurrent.

Proof: Note that the dual theorem coincides with the original theorem. So, if we can prove the 'if' part, then the dual theorem will automatically imply the 'only if' part. The 'if' part can be however proved by Menelaus Theorem.

Let lines $A_1A_2$, $B_2B_2$, $C_1C_2$ concur at $O$.Considering triangles $B1C1O$, $C1A1O$, and $A1B1O$, with the respective transversals $B2C2P$, $C2A2Q$, $A2B2R$, by Menelaus' theorem,

$$\frac{B_1P} {PC_1}·\frac{C_1C_2} {C_2O}·\frac{OB_2} { B_2B_1}=-1$$ $$\frac{C_1Q} {QA_1}·\frac{A_1A_2} {A_2O}·\frac{OC_2} { C_2C_1}=-1$$ $$\frac{A_1R} {RB_1}·\frac{B_1B_2} {B_2O}·\frac{OA_2} { A_2A_1}=-1$$

Multiplying the three equalities we obtain after simplification:

$$\frac{B_1P} {PC_1}·\frac{C_1Q} {QA_1}·\frac{A_1R} {RB_1}=-1$$

which by converse of Menelaus' theorem, implies the collinearity of the three points P, Q, R. Q.E.D