Geometry Parallelogram Proofs

Geometry proofs are basic laws of geometry figures, using these geometry proofs the geometry problems are solved. The Geometry parallelogram proofs are the proofs for parallelogram that help the rules of solving parallelogram based problems. A parallelogram proofs form an argument that establishes the truth of the parallelogram figure. Let us see basic geometry parallelogram proofs.

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Geometry Parallelogram Proofs:

Properties of parallelogram:

We know that a quadrilateral is a closed plane figure formed by four line segments and a parallelogram is a quadrilateral in which the opposite sides are parallel to each other.

Proof in geometry 1: In a parallelogram, the opposite sides are of equal length.

Parallelogram proof

Proof: Let ABCD be a parallelogram. Link up BD. take the triangles ABD and BDC.

Since AB || CD and BD is a transversal of AB and CD, m∠ABD = m∠BDC

Since AD|| BC and BD is a transversal of AD and BC, m∠ADB = m∠DBC

The side BD is common to both ABD and BDC.

Hence, by AAS property, ΔABD ≡ ΔBDC.

∴ The particular sides are equal.

AD = BC and AB = CD

Proof in geometry 2: In a parallelogram, the opposite angles are of equal measure.
Parallelogram proof
Proof: Let ABCD be the parallelogram. Link up BD.

Since AB || DC and BD is a transversal to AB and DC,
m∠ABD = m∠BDC
Since AD || BC and BD is a transversal
to AD and BC,
m∠ADB =m∠CBD
∴ m∠ABC = m∠ABD + m∠DBC
= m∠BDC+ m∠ADB
= m∠ADC
Similarly, m∠BAD=m∠BCD
Geometry Parallelogram Proofs:

Proof in geometry 3: The diagonals of a parallelogram bisect each other.
Parallelogram proof
Proof: ABCD is a parallelogram. AC and BD form diagonals.
By ASA criterion, ΔAMB ≡ CMD
∴AM = CM, BM = DM.
∴This shows diagonals bisect each other.


Proof in geometry 4: If the opposite sides of a quadrilateral are of equal length, then the quadrilateral is a parallelogram.
Parallelogram proof
Proof: Let ABCD be a quadrilateral where AD = BC,AB = CD. Link up AC.

Consider the triangles from the figure, ACB and ADC. By SSS criterion, ΔABC ≡ ΔCDA

Then m∠ BAC = m∠ACD,m∠CAD=m∠ACB
∴ AB || CD and AD || BC. Hence ABCD form a parallelogram.