Difference between revisions of "Rhombus"

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Corresponding parts of congruent triangles are congruent, so all 4 angles (the ones in the middle) are congruent.
 
Corresponding parts of congruent triangles are congruent, so all 4 angles (the ones in the middle) are congruent.
  
This leads to the fact that they are all equal to 90 degrees, and the diagonals are perpendiclar to eachother.
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This leads to the fact that they are all equal to 90 degrees, and the diagonals are perpendicular to each other.
  
 
== Example Problems ==
 
== Example Problems ==

Latest revision as of 12:20, 7 July 2016

A rhombus is a geometric figure that lies in a plane. It is defined as a quadrilateral all of whose sides are congruent. It is a special type of parallelogram, and its properties (aside from those properties of parallelograms) include:


Proofs

Proof that a rhombus is a parallelogram

All sides of a rhombus are congruent, so opposite sides are congruent, which is one of the properties of a parallelogram.

Or, there is always the longer way:

In rhombus $ABCD$, all 4 sides are congruent (definition of a rhombus).

$AB\cong CD$, $BC\cong DA$, and $AC\cong AC$.

By the SSS Postulate, $\triangle ABC\cong\triangle CDA$.

Corresponding parts of congruent triangles are congruent, so $\angle BAC\cong BCA$ and $\angle B\cong\angle D$. The same can be done for the two other angles, so $\angle A\cong\angle C$.

Convert the congruences into measures to get $m\angle A=m\angle C$ and $m\angle B=m\angle D$. Adding these two equations yields $m\angle A+m\angle B=m\angle C+m\angle D$.

The interior angles of a quadrilateral add up to 360 degrees, so $m\angle A+m\angle B+m\angle C+m\angle D=360$, or $m\angle A+m\angle B=360-m\angle C-m\angle D$.

Substituting gives $m\angle C+m\angle D=360-m\angle C-m\angle D$. When simplified, $m\angle C+m\angle D=180$.

If two lines are cut by a transversal and same-side interior angles add up to 180 degrees, the lines are parallel. This means $AD\|BC$. The same can be done for the other two sides, and know we know that opposite sides are parallel. Therefore, a rhombus is a parallelogram.

Proof that the diagonals of a rhombus divide it into 4 congruent triangles

In rhombus $ABCD$, $M$ is the point at which the diagonals intersect.

Since the diagonals of a rhombus are bisectors of eachother, $AM\cong MC$ and $BM\cong MD$.

Also, all sides are congruent.

By the SSS Postulate, the 4 triangles formed by the diagonals of a rhombus are congruent.

Proof that the diagonals of a rhombus are perpendicular

Continuation of above proof:

Corresponding parts of congruent triangles are congruent, so all 4 angles (the ones in the middle) are congruent.

This leads to the fact that they are all equal to 90 degrees, and the diagonals are perpendicular to each other.

Example Problems

Introductory

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