Properties of Quadrilateral

Statement: The line segments joining the ends towards the same side of two equal and parallel line segments are also equal and parallel to each other.

Given:

Two line segments AB and CD such that AB = CD and AB // CD. Points A,C and B,D are joined.

To prove:

AC = BD and AC // BD.

Construction: (Click me)

Proof: (Click me)

S. No.	Statements	S. No.	Reasons
1.	In ΔABC and ΔBCD	1.
(i)	AB = CD (S)	(i)	Given
(ii)	∠ABC = ∠BCD (A)	(ii)	Alternate angles AB // CD
(iii)	BC = BC (S)	(iii)	Common side
2.	ΔABC ≅ ΔBCD	2.	SAS axiom
3.	AC = BD	3.	Corresponding sides of congruent triangles
4.	∠ACB = ∠CBD	4.	Corresponding angles of congruent triangles are equal
5.	AC // BD	5.	Alternate angles
6.	AC = BD and AC // BD	6.	From statements 3 and 5

Proved.

Statement: The line segments joining the opposite ends of two equal and parallel line segments bisect each other.

Given:

In quadrilateral PQRS, PQ = RS and PQ || RS. PS and QR are diagonals intersecting at O.

To prove:

PO = OS and QO = OR.

Construction: (Click me)

Proof: (Click me)

S. No.	Statements	S. No.	Reasons
1.	In ΔPOQ and ΔROS	1.
(i)	∠OPQ = ∠OSR (A)	(i)	Alternate angles PQ//RS
(ii)	PQ = RS (S)	(ii)	Given
(iii)	∠PQO = ∠SRO (A)	(iii)	Alternate angles PQ//RS
2.	ΔPOQ ≅ ΔROS	2.	ASA axiom
3.	PO = OS, QO = OR	3.	Corresponding sides of congruent triangles

Proved.

Statement: In a parallelogram, opposite sides are equal and opposite angles are equal.

Given:

A parallelogram ABCD where AB || DC and AD || BC.

To prove:

AB = DC, AD = BC and ∠ABC = ∠ADC, ∠DAB = ∠BCD

Construction: (Click me)

Proof: (Click me)

S. No.	Statements	S. No.	Reasons
1.	In ΔABD and ΔBCD	1.
(i)	∠ABD = ∠BDC (A)	(i)	AB//DC Alternate angles
(ii)	BD = BD (S)	(ii)	Common sides
(iii)	∠ADB = ∠DBC (A)	(iii)	AD//BC Alternate angles
2.	ΔABD ≅ ΔBCD	2.	ASA axiom
3.	AB = DC and AD = BC	3.	Corresponding sides of congruent triangles
4.	∠DAB = ∠BCD ∠ABD = ∠BDC ∠ADB = ∠DBC	4.	Corresponding angles of congruent triangles
5.	∠ABD + ∠DBC = ∠BDC + ∠ADB	5.	Equal axiom
6.	∠ABC = ∠ADC	6.	From statement 5
7.	AB = DC, AD = BC and ∠ABC = ∠ADC, ∠DAB = ∠BCD	7.	From statements 3, 4 and 6

Proved.

Statement: A quadrilateral having opposite sides equal is a parallelogram.

Given:

A quadrilateral ABCD where AB = CD and AD = BC.

To prove:

ABCD is a parallelogram.

Construction: (Click me)

Proof: (Click me)

S. No.	Statements	S. No.	Reasons
1.	In ΔABC and ΔACD	1.
(i)	AB = CD (S)	(i)	Given
(ii)	AD = BC (S)	(ii)	Given
(iii)	AC = AC (S)	(iii)	Common side
2.	ΔABC ≅ ΔACD	2.	SSS axial
3.	∠ACB = ∠DAC ∠BAC = ∠ACD	3.	Corresponding angles of congruent triangles
4.	AB \|\| CD , AD \|\| BC	4.	Being alternate angles

Proved.

Statement: A quadrilateral with equal opposite angles is a parallelogram.

Given:

Quadrilateral ABCD where ∠ABC = ∠CDA and ∠DAB = ∠BCD.

To prove:

ABCD is a parallelogram.

Proof: (Click me)

S. No.	Statements	S. No.	Reasons
1.	∠ABC + ∠BCD + ∠CDA + ∠DAB = 360°	1.	The sum of interior angles of quadrilateral
2.	∠ABC + ∠BCD + ∠ABC + ∠BCD = 360° or, 2∠ABC + 2∠BCD = 360° or, ∠ABC + ∠BCD = 180°	2.	Being ∠ABC = ∠CDA and ∠DAB = ∠BCD
3.	AB // CD	3.	Being sum of co-interior angles
4.	Similarly ∠BCD + ∠CDA = 180°	4.	Reasons like: 1, 2, 3
5.	BC // AD	5.	Being sum of co-interior angles
6.	AB // CD , BC // AD	6.	From statements 3 and 5
7.	ABCD is a parallelogram	7.	From statement 6

Statement: Diagonals of a parallelogram bisect each other.

Given:

Parallelogram PQRS where diagonals PR and QS intersect at O.

To prove:

PO = OR and QO = OS.

Proof: (Click me)

S. No.	Statements	S. No.	Reasons
1.	In ΔPOQ and ΔROS	1.
(i)	∠OPQ = ∠ORS (A)	(i)	Alternate angles PQ // SR
(ii)	PQ = RS (S)	(ii)	Opposite sides of parallelogram
(iii)	∠OQP = ∠OSR (A)	(iii)	Alternate angles PQ // SR
2.	ΔPOQ ≅ ΔROS	2.	ASA axiom
3.	PO = OR, QO = OS	3.	Corresponding sides of congruent triangles

PROVED

Statement: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Given:

Quadrilateral ABCD in which diagonals AC and BD intersect at O such that AO = OC and OB = OD.

To prove:

ABCD is a parallelogram.

Proof: (Click me)

S. No.	Statements	S. No.	Reasons
1.	In ΔAOB and ΔDOC	1.
(i)	AO = OC (S)	(i)	Given
(ii)	∠AOB = ∠COD (A)	(ii)	Vertically opposite angles
(iii)	OB = OD (S)	(iii)	Given
2.	ΔAOB ≅ ΔDOC	2.	SAS axiom
3.	AB = DC	3.	Corresponding sides of congruent triangles
4.	∠OBA = ∠ODC	4.	Corresponding angles of congruent triangles
5.	AB // DC	5.	Alternate angles
6.	AD // BC, AD = BC	6.	AB // DC and AB = DC
7.	ABCD is a parallelogram	7.	Opposite sides are equal and parallel

PROVED