Parallelogram Theorem

Quadrilateral means a closed figure formed by four line segments and a parallelogram is a quadrilateral in which the opposite sides are parallel to each other. A point is used to represent a position in space. A plane means a surface increasing infinitely in every directions such that all points lying on the line joining any two points on the surface. The parallelogram theorems are given below.

Parallelogram Proof

Theorem 1:

Parallelograms are the same base and between the same parallel lines are equal in area.
Activity:
Draw a line segment AB. Draw a line l parallel to AB. Mark a point C on l. Draw AL perpendicular to l. Measure the length of AL.


We find the area of the triangle ABC as 1/2 × base × height
= 1/2 * AB * AL
Mark another point P on l. We find the area of ?ABP as 1/2 × base × height = 1/2 * Base * Height
= 1/2 * AB * AL
thus, we seen that the area of the triangle remains the same for all positions of the vertex C on the line


Theorem2:

A parallelogram is a rhombus if its diagonals are 90 degree.
Given: ABCD is a parallelogram where the diagonals AC and BD are perpendicular.
To prove: ABCD is a rhombus.
Construction: Draw the diagonals AC and BD. Let M represent the point of intersection of AC and BD (see Figure).
Proof: In triangles AMB and BMC,
(i) ?AMB = ?BMC = 90°
(ii) AM = MC
(iii) BM is common.
? By SSA criterion, ?AMB = ?BMC.
In particular, AB = BC.
Since ABCD is a parallelogram, AB = CD, BC = AD.
? AB = BC = CD = AD.
Hence ABCD is a rhombus. The theorem is proved.

Theorem 3:

The quadrilateral is a parallelogram, then the one pair of opposite sides are parallel and equal.

Given: ABCD is a quadrilateral, where AB || CD and AB = CD.
To prove: ABCD is a parallelogram.
Construction: Draw the diagonal AC (see Figure).
Proof: In triangles ABC and ADC,
(i) AB = CD (given)
(ii) AC is common
(iii) m?BAC = m?ACD
By SAS criterion, ?ABC = ?ADC. sides are equal
AD = BC, m?DAC = m?ACB.
AD || BC.
Hence ABCD is a parallelogram. The theorem is proved

Theorem:
The diagonals of a parallelogram bisect each other.
Proof:
ABCD is a parallelogram. AC and BD are diagonals.
By ASA criterion, ?AMB = CMD (see Figure).
? AM = CM, BM = DM.
? The diagonals bisect each other.

Eucledian Proof

 

According to scientist Adler, in his book on modern geometry he mention “One of the main defects in traditional Euclidean proof is its almost complete disregard of such notions as the two sides of a line and the interior of an angle. Without clarification of these ideas, absurd consequences result.”
Here is an illustration to show traditional proof of a geometrical figure parallelogram PQRS

Let PQRS parallelogram where, PQ||RS and QR||PS and O be the intersection of the diagonals
PR and QS. Prove PO = RO
The traditional proof (Euclidean Proof):
To prove :
ΔPRQ = RSP (hence PQ = RS) and ΔPOQ = ΔSOR (hence PO = RO)
Proof :
By congruency of triangle ∠RPQ = ∠PRS
because alternative angles of parallels PO and RS.
As we assumed the trivial fact that point S and Q are either sides of the line PR .
Draw back lies in harder situation to prove the last point.