In the figure, l ïï m and M is the mid-point of AB. Prove that M is also the mid-point of CD.  

Since l ïï m, and AB is the transversal 
∴ 탊M = ∠DBM ...... (Alternate interior angles)
Now in ΔACM and ΔBDM, we have
∴  ∠CAM = ∠DBM ..... (proved above)
    AM = BM ...... (M is the mid-point)
ഊMC = ∠BMD ............(Vertically opposite angles)
∴  ΔACM ≅ ΔBDM
 ∴ DM = MC .............(CPCT)
Hence, M is the mid-point of DC.