Answers

2014-10-27T19:42:54+05:30
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  

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2014-10-27T19:43:20+05:30
 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
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