Question

Prove that the diagonals of a rhombus are perpendicular bisectors of each other.​

Answer 1

$\sf ANSWER$

$\textbf{Theorem :}$

Diagonals of a rhombus are perpendicular bisectors of each other.

$\textbf{Given :}$

□EFGH is a rhombus.

$\textbf{To\: prove :}$

Diagonal EG is the perpendicular bisectors of diagonal HF.

Diagonal HF is the perpendicular bisectors of diagnal EG.

$\textbf{prove :}$

(i) seg EF = seg EH ( Given )

seg GF = seg GH ( given )

Every points which is equidistant from the end points if a segment is on the perpendicular bisector of the segment .

Therefore, point E and point G are on the perpendicular bisectors of seg HF.

one and only one line passes through two distinct points.

Therefore, Line EG is the perpendicular bisectors of diagonal HF.

Therefore, diagonal EG is the perpendicular bisector of the diagonal HF.

(ii) Similarly, we can prove that diagonal HF is the perpendicular bisector of EG.

More Note !

diagonals of a rectangle are congruent .

Diagonals of a rhombus are perpendicular bisector of each other.

diagonals of a square bisect opposite angles .

diagonal of a square are congruent

diagonal of a Rhombus bisect the pairs of opposite angles .

diagonals of a square are perpendicular bisectors of each other .