Given: Square TDZI us rhombus Prove: Triangle DEZ is congruent to Triangle TEI

Question

I need help with this.

Answer 1

Assuming that TDZI is both a rhombus and a square, and that point E is where the diagonals bisect each other.

Assuming that the top horizontal segment from left to right is TD, and the bottom horizontal segment from left to right is IZ

 

Segment TI and DZ are congruent to each other (properties of a square/rhombus: opposite sides are congruent)

Segment IE and ED are congruent to each other (diagonals of a square/rhombus bisect each other)

Segment TE and TZ are congruent to each other (diagonals of a square/rhombus bisect each other)

Triangle DEZ is congruent to Triangle TEI (SSS)

 

You could also just do the two diagonals bisecting each other and that angle TEI is congruent to angle DEZ (vertical angles).  Then you can conclude that the triangles are congruent by SAS.