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Special Parallelogram
Definition of Rectangles
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.
Rectangle Property 1:
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}
Rectangle Property 2:
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.
Corollary of rectangle property
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
Definition of rhombus
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.
Rhombus property 1:
A rhombus is a parallelogram having four congruent sides.
In the figure above, if ABCD is a rhombus, then
AB = BC = CD = DA
Rhombus property 2:
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
angle 7 = angle 8
Rhombus property 3:
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
Area of rhombus
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
Definition of square
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.
Property of 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.