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Show that diagonals of square divide it into two congruent triangles

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Given: A square ABCDTo prove: ∆BAD ≅ ∆BCD, ∆ABC ≅ ∆ADCConstruction: Join BDProof: In ∆BAD and ∆BCD,AB = CD                          [Opposite sides of a square are equal]AD = BC                          [Opposite sides of a square are equal]BD = BD                          [Common]∴ By SSS congruence criterion,∆BAD ≅ ∆BCDSimilarly, ∆ABC ≅ ∆ADC.Hence, the diagonals of a square divide it into two congruent triangles.
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as the angles are 90 degree and all the sides are equal,it is possible
 
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Let ABCD be a square and line AC be its diagonal. We know, in a square all the sides are equal and all the angles are right angled.?
So, in ABC and ACD,?
AB=CD
angle ABC =angle ADC?
AD =BC
So, by SAS congruency both the triangles are equal.
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