Given ABCD is a rhombus and DF = BE. Prove that triangle \[\vartriangle ABE \cong \vartriangle DFC\]
 

Hint: In a triangle if any two sides and the angle in between those two sides are equivalent to the corresponding two sides and the angle in between those two sides of any other triangle then according to side angle side rule those two triangles are equivalent to each other.
Given: ABCD is a rhombus and DF = BE
To prove: triangle ABE is equivalent to triangle DFC

Complete step-by-step solution:
As we know, a rhombus is a special case of parallelogram which is formed by four sides and in which all four sides are equal in length and also in rhombus two opposite angles are equal to each other.
So here it is given that ABCD is a rhombus so according to the properties of rhombus we can say that,
\[AB = CD\]
\[DA = BC\]
\[\angle ABC = \angle CDA\]and \[\angle DAB = \angle BCD\]
Now comparing triangle ABE and triangle DFC
According to the rhombus properties, we have
\[AB = CD\] (all four sides are equal in length) (1)
Angle ABE = angle CDF (two opposite angles are equal to each other) (2)
Now in the question it is given that,
\[DF = BE\] (3)
Now from congruence of triangles we know that two triangles are equivalent to each other if their two sides and angle between the two sides are equivalent to each other.
Therefore, from (1), (2) and (3) it is clear that SAS rule is satisfied in \[\vartriangle ABE\]and\[\vartriangle DFC\]. Hence from SAS rule it is proved that \[\vartriangle ABE \cong \vartriangle DFC\]

Note: There are other properties of congruence of triangle like
SSS (Side-Side-Side)
SAS (Side-Angle-Side)
ASA (Angle-Side-Angle)
AAS (Angle-Angle-Side)
RHS (Right angle-Hypotenuse-Side)
Here in the congruence of triangles we find the similarity in sides and angles.