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# Special Parallelogram

### Definition of Rectangles

### Rectangle Property 1

### Rectangle Property 2

### Corollary of rectangle property

### Definition of rhombus

### Rhombus property 1

### Rhombus property 2

### Rhombus property 3

### Area of rhombus

### Definition of square

### Property of square

In a parallelogram, if one of the angles is a right angle, then the parallelogram is a rectangle.

In the figure above, if ABCD is a parallelogram and angle A = 90^{o}, then ABCD is a rectangle.

Each angle of the four angles of a rectangle is 90 degrees.

In the figure above, if ABCD is a rectangle, then angle A = angle B = angle C = angle D = 90^{o}.

The diagonals of a rectangle are congruent.

In the figure above, if ABCD is a rectangle, AC and BD are two diagonals, then AC = BD.

The length of the median of hypotenuse of a right triangle equals to one-half the length of the hypotenuse.

In the figure above, if ABC is a right triangle, AB is the hypotenuse, D is the midpoint of AB, then CD = (1/2)AB.

A rhombus is a parallelogram having pair of congruent consecutive sides.

In the figure above, if ABCD is a parallelogram, and AD = AB, then ABCD is a rhombus.

A rhombus is a parallelogram having four congruent sides.

In the figure above, if ABCD is a rhombus, then AB = BC = CD = DA.

The diagonals of a rhombus bisect the angles at the vertices which they join.

In the figure above, if ABCD is a rhombus, then angle 1 = angle 2, angle 3 = angle 4, angle 5 = angle 6 and angle 7 = angle 8.

The diagonals of a rhombus is perpendicular to each other.

In the figure above, if ABCD is a rhombus, and AC and BD are two diagonal, then AC is perpendicular to DB.

If the lengths of diagonals of a rhombus are a and b respectively, then its area is one-half the product of a and b.

In the figure above, if ABCD is a rhombus, the length of AC is a and the length of DB is b , then the area of the rhombus = (1/2) a b.

A square is a rectangle having four congruent sides.

In the figure above, if ABCD is a rectangle and AB = BC = CD = DA, then ABCD is a square.

A square have four right angles, four congruent sides, two congruent diagonals, diagonals bisect opposite angles, diagonals are perpendicular one another.

In the figure above, if ABCD is a square, then 1. angle A = angle B = angle C = angle D = 90^{o} (four right angles). 2. AB = BC = CD = DA (four congruent sides).
3. AC = DB (two congruent diagonals). 4. angle 1 = angle 2 = 45^{o} (diagonals bisect opposite angles). 5. AC is perpendicular to BD (diagonals are perpendicular one another).

**Example 1**

In the figure below, ABCD is a rectangle. E is the midpoint of AB. Prove ED = EC.

- Proof
- Since ABCD is a rectangle (given), so
- AD = BC (opposite sides of a rectangle are congruent)
- angle A = angle B = 90
^{o}(each angle of the four angles of a rectangle is 90 degrees) - Since E is the midpoint of AB (given), so AE = BE (property of midpoint)
- In triangle DAE and triangle CBE,
- since AD = BC, angle A = angle B, AE = BE, so
- triangle DAE and triangle CBE are congruent (SAS postulate)
- so ED = EC (In congruent triangles, congruent angles opposite congruent sides.)

**Example 2**

In the figure below, ABCD is a rectangle, AE = BF. Prove EC = FD.

- Proof
- Since ABCD is a rectangle (given), so
- AD = BC (opposite sides of a rectangle are congruent)
- angle A = angle B (each angle of the four angles of a rectangle is 90 degrees)
- Since AE = BF (given), so
- AB - BF = AB - AE
- note: AB - BF = AF and AB - AE = BE
- so AF = BE
- In triangle ADF and triangle BCE,
- since AD = BC, angle A = angle B, AF = BE, so
- triangle ADF and triangle BCE are congruent (SAS postulate)
- so FD = EC

**Example 3**

In the figure below, ABCD is a rhombus. E is the midpoint od AD. F is the midpoint of DC. If EC = FB, prove ABCD is a square.

- Proof
- Since ABCD is a rhombus (given),
- so AB = BC = CD = DA (A rhombus is a parallelogram having four congruent sides.)
- and angle A = angle C, angle B = angle D (A rhombus is a parallelogram having congruent opposite angles.)
- Since E id the midpoint of AD (given), so DE = AE (Property of the midpoint)
- Since F is the midpoint of DC (given), so CF = FD (Property of the midpoint)
- so DE = AE = CF = FD = (1/2)AB
- In triangle DEC and triangle CFB,
- since DE = CF, EC = FB (given), CD = CD = BC,
- so triangle DEC and triangle CFB are congruent (SSS postulate)
- so angle D = angle BCF (in congruent triangles, congruent sides opposite congruent angles.)
- Since point F lies on CD, so angle BCF = angle BCD
- Since angle D + angle BCD = 180
^{o}(Property of a rhombus), - so angle D = angle BCD = 90
^{o} - so ABCD is a square.