Question

Prove that all 4 sides of rhombus are equal

Answer 1

You can use the following six methods to prove that a quadrilateral is a rhombus. The last three methods in this list require that you first show (or be given) that the quadrilateral in question is a parallelogram:

If all sides of a quadrilateral are congruent, then it’s a rhombus (reverse of the definition).

If the diagonals of a quadrilateral bisect all the angles, then it’s a rhombus (converse of a property).

If the diagonals of a quadrilateral are perpendicular bisectors of each other, then it’s a rhombus (converse of a property).

Tip: To visualize this one, take two pens or pencils of different lengths and make them cross each other at right angles and at their midpoints. Their four ends must form a diamond shape — a rhombus.

If two consecutive sides of a parallelogram are congruent, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).

If either diagonal of a parallelogram bisects two angles, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).

If the diagonals of a parallelogram are perpendicular, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).

Here’s a rhombus proof for you. Try to come up with a game plan before reading the two-column proof.

Answer 2

All 4 sides of rhombus are equal.

• Given: Let ABCD be a rohmbus.

• To prove: AB=BC=CD=AD

• Proof: In ∆ABC and ∆ACD,

angleACB=angleCAD and

angleACD=angleCAB (Alternate interior angle)

AC=AC (common)

• Therefore, by ASA congruence

∆ABC is congruent to∆ACD

• So, AB=CD and BC=AD _(i)

• We know that the diagonal of rhombus bisects each other at 90°.

•Let AC and BD bisect at O.

• In ∆AOB and ∆BOC,

angleAOB=angleBOC=90°

AO=OC

BO=BO (common)

• Therefore, by SAS congruence

∆AOB is congruent to ∆BOC.

• So, AB=BC _(ii)

• From (i) and (ii),

AB=BC=CD=AD

• Hence, all 4 sides of rhombus are equal.