Question

Given ABCD is a rhombus and DF=BE. Prove that triangle ABE is congruent to triangle DFC.

Answer 1

<B = <D (opposite sides of parallelogram)

180 - <B = 180 - <D

or < CBE = <FDC  -------------(1)

AB = DC (opposite sides of a parallelogram)

also, AB = BE (given)

so, BE = DC ----------(2)

BC = AD (opposite sides of a parallelogram)

also, AD = DF (given)

so, BC = DF -------(3)

Now, in triangle BEC and DCF,

BE = DC (from 2)

< CBE = <FDC (from 1)

BC = DF (from 3)

so, triangle BEC congruent triangle DCF (by SAS postulate)

Hope that this answer will help you ✌️

Answer 2

DF=BE(Given)

DC=AB(all the sides of rhombuses are equal)

Angle D =Angle B(opposite angles of rhombuses are equal)

:. ∆ABE≈∆DFC