Question

show that the diagonals of rhombus divide it into four congruent triangles

Answer 1

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Answer 2

Let the rhombus be ABCD and let there diameters intersect each other at O.
Since,in a Rhombus all sides are equal,
Therefore AB=AD
and since ABCD is rhombus therefore its diagonals bisect each other
So,AO=DO
and
AO=AO(common)
Therefore 
ΔABO=~ΔADO
Similarly,ΔABO=~ΔBCO
Similarly,
ΔBCO=~ΔDCO
Therefore 
ΔABO=~ΔBCO=~ΔCDO=~ΔADO