A quadrilateral is a foursided 2D figure. By showing that quadrilaterals have specific properties, we can narrow it down to the most specific type of quadrilateral.
We define a parallelogram as any quadrilateral whose opposite side lengths are parallel. In some instances, a parallelogram can also be a rhombus, rectangle or square, depending on what knowledge we know about the parallelogram.
Properties of parallelograms:
We define a rhombus as a parallelogram whose all four sides are congruent.
Rhombus 
Properties of a rhombus:
We define a rectangle as a parallelogram whose interior angles are right angles.
Rectangle 
Properties of rectangles:
And lastly, we define a square as a parallelogram whose all four sides are congruent and and all interior angles are right angles.
Square 
Properties of squares:
If we can prove that a parallelogram possesses one of these necessary conditions, then we can prove that it is a rhombus, rectangle or square respectively.
Consider the following theorem.
If a parallelogram has one pair of consecutive sides that are congruent, then the parallelogram is a rhombus.
To begin proving the above theorem, we can first label the parallelogram by its set of vertices, $PQRS$PQRS.
Parallelogram $PQRS$PQRS 
From the given information, we know that a pair of consecutive sides are congruent. So let's choose $\overline{PQ}$PQ and $\overline{SP}$SP. Ultimately we could have chosen another pair, but the proof would be the same. We write this as a geometric statement $\overline{PQ}\cong\overline{SP}$PQ≅SP.
Parallelogram $PQRS$PQRS with congruent sides 
We also know that the $PQRS$PQRS is a parallelogram, and one feature of the parallelogram is that opposite sides are congruent. We can state this as $\overline{PQ}\cong\overline{RS}$PQ≅RS and $\overline{SP}\cong\overline{QR}$SP≅QR. Congruence is transitive, meaning that if two things are congruent to the same thing, they are also congruent. This means that $\overline{PQ}\cong\overline{QR}$PQ≅QR and $\overline{SP}\cong\overline{RS}$SP≅RS and more generally all the sides are congruent. By definition, a rhombus is a parallelogram with all sides congruent. So $PQRS$PQRS is a rhombus and we are finished with the proof.
We can formalize the above steps into a twocolumn proof where each line contains a geometric statement in the left column and a corresponding reason in the right column.
Given the parallelogram $PQRS$PQRS, and the fact that $\overline{PQ}\cong\overline{SP}$PQ≅SP, prove that $PQRS$PQRS is a rhombus. 

Statements  Reasons 
$PQRS$PQRS is a parallelogram and $\overline{PQ}\cong\overline{SP}$PQ≅SP  Given 
$\overline{PQ}\cong\overline{RS}$PQ≅RS and $\overline{SP}\cong\overline{QR}$SP≅QR 
If a quadrilateral is a parallelogram, then its opposite sides are congruent. 
$\overline{PQ}\cong\overline{QR}\cong\overline{RS}\cong\overline{SP}$PQ≅QR≅RS≅SP 
Transitive property of congruence 
$PQRS$PQRS is a rhombus  Definition of a rhombus 
The final line contains the statement that the quadrilateral is a rhombus, which is what we wanted to prove.
In order to show that a given parallelogram is a rectangle, which of the following must be proved?
The parallelogram has four congruent sides.
The parallelogram has four right angles.
The parallelogram has opposite sides congruent.
The parallelogram has opposite sides parallel.
The parallelogram has opposite angles congruent.
The parallelogram has four congruent sides.
The parallelogram has four right angles.
The parallelogram has opposite sides congruent.
The parallelogram has opposite sides parallel.
The parallelogram has opposite angles congruent.
Given that $PQRS$PQRS is a parallelogram, and $\overline{PR}$PR and $\overline{QS}$QS are perpendicular, prove that $PQRS$PQRS is a rhombus.
We define a trapezoid as any quadrilateral with one pair of opposite sides that are parallel.
A special type of a trapezoid, called an isosceles trapezoid, occurs when the two legs of the trapezoid are congruent and hence the diagonals are congruent.
Isosceles trapezoid 
There are a number of theorems that explain equivalent ways of identifying whether a trapezoid is isosceles. We call these special theorems criteria.
Trapezoid with base angles congruent 
Trapezoid with $\angle1$∠1 and $\angle3$∠3 are supplementary, and $\angle2$∠2 and $\angle4$∠4 are supplementary 
If a trapezoid has congruent diagonals, then it is an isosceles trapezoid.
Trapezoid with congruent diagonals 
Consider the following theorem.
If a trapezoid has congruent base angles, then it is an isosceles trapezoid.
To begin proving the above theorem, we can first label the trapezoid by its set of vertices, $ABCD$ABCD. Let $\angle ADC$∠ADC and $\angle BCD$∠BCD be congruent.
Trapezoid $ABCD$ABCD 
We next construct the pair of altitudes $\overline{AP}$AP and $\overline{BQ}$BQ, which form the pair of triangles $\triangle APD$△APD and $\triangle BQC$△BQC. If we can show that these triangles are congruent, then we can show that $\overline{AD}$AD is congruent to $\overline{BC}$BC, which will mean that $ABCD$ABCD is an isosceles trapezoid.
Trapezoid $ABCD$ABCD with altitudes drawn 
Firstly we are given that $\angle ADC$∠ADC and $\angle BCD$∠BCD are congruent. Since altitudes create right angles with the base of a trapezoid, $\angle APD$∠APD and $\angle BQC$∠BQC are right angles. All right angles are congruent, so we have that $\angle APD$∠APD and $\angle BQC$∠BQC are congruent. The altitudes themselves are congruent, so $\overline{AP}$AP is congruent to $\overline{BQ}$BQ. This means that $\triangle APD\cong\triangle BQC$△APD≅△BQC are congruent by angleangleside congruence theorem.
It follows that the two sides $\overline{AD}$AD and $\overline{BC}$BC are congruent because they are corresponding parts of the two congruent triangles. Hence the trapezoid $ABCD$ABCD is isosceles, by definition.
We can formalize the above steps into a twocolumn proof where each line contains a geometric statement in the left column and a corresponding reason in the right column.
Statements  Reasons 
$ABCD$ABCD is a trapezoid, $\overline{AP}$AP and $\overline{BQ}$BQ are altitudes, and $\angle ADC\cong\angle BCD$∠ADC≅∠BCD 
Given 
$\angle APD$∠APD and $\angle BQC$∠BQC are right angles  Definition of an altitude 
$\angle APD\cong\angle BQC$∠APD≅∠BQC  All right angles are congruent. 
$\overline{AP}\cong\overline{BQ}$AP≅BQ  Altitudes of a trapezoid are congruent 
$\triangle APD\cong\triangle BQC$△APD≅△BQC 
Angleangleside congruence theorem 
$\overline{AD}\cong\overline{BC}$AD≅BC  Corresponding parts of congruent triangles are congruent (CPCTC) 
$ABCD$ABCD is an isosceles trapezoid  Definition of an isosceles trapezoid 
Below is a summary of the criteria for a trapezoid to be an isosceles trapezoid.
A trapezoid is an isosceles trapezoid if any of the following are true:
In order to show that a given trapezoid is isosceles, which of the following must be proved?
The trapezoid has diagonals that bisect each other.
The trapezoid has diagonals that are perpendicular.
The trapezoid has diagonals that are congruent.
The trapezoid has diagonals that are perpendicular bisectors of each other.
The trapezoid has diagonals that bisect each other.
The trapezoid has diagonals that are perpendicular.
The trapezoid has diagonals that are congruent.
The trapezoid has diagonals that are perpendicular bisectors of each other.
Given that $ABCD$ABCD is a trapezoid with altitudes $\overline{AP}$AP and $\overline{BQ}$BQ and that $\angle ADC$∠ADC is congruent to $\angle BCD$∠BCD, prove that $ABCD$ABCD is an isosceles trapezoid.
Prove and apply theorems about parallelograms.
Use congruence and similarity criteria for triangles to solve problems algebraically and geometrically.
Use congruence and similarity criteria for triangles to prove relationships in geometric figures.