Question

In the adjoining figure,ABCD is a parallelogram. AC its Diagonal. Prove that AC divides parallelogram ABCD into two congruent triangles 

Answer 1

Consider two triangles ΔABC and ΔADC (fig is attached to this)

AC = AC (common)
AD = BC ( property of a parallelogram)
DC = AB ( property of a parallelogram)

So ΔABC ≈ Δ ADC ( SSS congruence)

Thus proved  

Answer 2

 In parallelogram AB=DC
AD=BC
and AC is cutting parallelogram into two triangle
In ΔABC and in ΔADC
    AB=BC
   BC=AD
   One side is common AC by SSS
ΔABC≡ΔADC