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### Parallelogram and its Properties

Definition of Parallelogram
A parallelogram is a quadrilateral having two pairs of parallel sides. In the figure above,
if AB // DC and AD // BC, then
ABCD is a parallelogtam.
Parallelogram Property 1
Opposite angles of a parallelogram are congruent. In the figure above,
if ABCD is a parallelogram, then
angle A = angle C and
angle B = angle D
note: angle A and angle C are a pair of opposite angles. Angle B and angle D are a pair of opposite angles.
Parallelogram Property 2
Opposites sides of a parallelogram are congruent. In the figure above,
if ABCD is a parallelogram, than
AB = DC and AD = BC
note: AB and DC are a pair of opposite sides. AD and BC are a pair of opposides.
Parallelogram Property 3
The diagonals of a parallelogram bisect each other. In the figure above,
if ABCD is a parallelogram, AC and BD are its two diagonals, then
AE = EC and BE = ED
note: the two diagonal AC and BD intersect at point E.
Parallelogram Property 4
Consecutive pairs of angles of a parallelogram are supplementary. In the figure above,
if ABCD is a parallelogram then
angle D + angle A = 180o
angle A + angle B = 180o
note: in parallelogram ABCD, angle D and angle A are a pair of consecutive angles and angle A = angle C, angle B = angle D.
Congruent lines
The parallels between two parallels are congruent. In the figure above,
if l1 // l2 and AB // CD // EF, then
AB = CD = EF
Distance between two parallels
Distance between two parallels is the line drawn from any point of a parallel to the distance of another parallel. In the figure above, if l1 // l2 , P is a point lies on l1 , Q is a point lies on l2 and PQ is perpendicular to l2, then
PQ is the distance between parallel lines l1 and l2.
Proving parallelogram
If both pairs of opposite sides of a quadrilateral are equal, then it is a parallelogram. In the figure above, if AB = DC and AD = BC, then
ABCD is a parallelogram.
note: AB and DC are a pair of opposite sides. AD and BC are a pair of opposite sides.
Proving parallelogram
If both pairs of opposite angles of a quadrilateral are equal, then it is a parallelogram. In the figure above, if angle A = angle C and angle B = angle D, then
ABCD is a parallelogram.
note: angle A and angle C are a pair of opposite angles. Angle B and angle D are a pair of opposite angles.
Proving parallelogram
If a pair of opposite sides of a quadrilateral are equal and parallel, then it is a parallelogram. In the figure above, if AB = DC and AB // DC, then
ABCD is a parallelogram.
note: AB and DC are a pair of opposite sides.
Proving parallelogram
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. In the figure above, AC and BD are two diagonals which intersect at point E,
if AE = EC and BE = ED, then
ABCD is a parallelogram.
Example 1:
In the figure below, ABCD is a parallelogram. If angle A = 120o, what is the degree measure of the angle D? Solution:
Since ABCD is a parallelogram (Given), so
angle A + angle D = 180o
(Consecutive pairs of angles of a parallelogram are supplementary.)
angle D = 180o - angle A
= 180o - 120o
= 60o
Therefore, the degree measure of the angle D is 120o
Example 2:
In the figure below, ABCD is a parallelogram. DB is a diagonal. AE is perpendicular to DB at E and CF is perpendicular to DB at F. Prove AE = CF. Proof:
Since ABCD is a parallelogram (given), so AB = DC and AB // DC (a pair of opposite sides of a parallelogram are equal and parallel)
Since AB // DC, so angle ABD = angle CDB (alternate interior angles are congruent)
Since AE is perpendicular to BD and CF is perpendicular to BD (given), so angle AEB = angle CFD = 90o (property of perpendicular lines)
Since point E lies on BD and point F lies on BD, so angle ABD = angle ABE, angle CDB = angle CDF
In triangle ABE and triangle CDF,
since angle AEB = angle CFD, angle ABE = angle CDF, AB = CD,
then triangle ABE is congruent to triangle CDF (AAS theorem)
so AE = CF (In two congruent triangles, congruent angles opposite congruent sides.)
Example 3 :
In the figure below, ABCD is a parallelogram. AC is a diagonal. If DE = BF, prove EG = FG. Proof:
Since ABCD is a parallelogram, so AB = DC and AB // DC (a pair of opposite sides of a parallelogram are equal and parallel)
Since BF = DE (given), so AB - BF = DC - DE, that is, AF = CE
Since AB // DC, so angle BAC = angle DCA and angle AFE = angle CEF (alternate interior angles are congruent)
Since point F lies on AB and point E lies on CD, AC and EF intersect at point G, so angle BAC = angle FAG and angle AFE = angle AFG. Same reason, angle CEF = angle CEG and angle ECA = angle ECG.
In triangle AFG and triangle CEG,
since angle FAG = angle EGC, AF = CE, angle AFG = angle CEG,
then triangle AFG is congruent to triangle CEG (ASA postulate)
so FG = EG (In two congruent triangles, congruent angles opposite congruent sides)