Theorem Of Parallelograms

Proofs of theorems related to Parallelograms.

Theorem 1

Theorem 1

A diagonal of a parallelogram divides it into two congruent triangles.


Given.
A parallelogram ABCD and diagonal AC divides it into two triangles △ABC and △CDA.

To prove. △ABC△CDA

Proof.
Statements
Reasons
In△ABC and △CDA
1. ∠BAC=∠ACD 1.Alt.∠s,AB∠DC and AC is transversal.
2.∠BCA=∠CAD 2. Alternate ∠s, BC∠AD and AC is transversal.
3. AC=CA
3. Common
4.△ABC≌△CDA 4. ASA congruency rule




Theorem 2

In a parallelogram, opposite sides are equal.

Given:
A parallelogram ABCD.

To Prove: AB=DC and BC=AD

Construction: Join AC.


Proof:

Statements
Reasons
In △ABC and △ CDA
1. ∠BAC=∠ACD 1. Alternate ∠s, AB ∥ DC and AC is transversal
2. ∠BCA=∠CAD 2. Alt, ∠s, BC∥AD and AC is transversal.
3. AC=CA
3. Common
4. △ABC △CDA 4. ASA Congruency.
5. AB= DC and BC=AD
5. c.p.c.t.

The converse of the above theorem is also true. i.e. In a quadrilateral if opposite sides are equal, it is a parallelogram.

Theorem 3

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Given. A quadrilateral ABCD in which AB = DC and BC = AD.
Construction. Join AC.

Proof:

Statements
Reasons
In△ABC and △CDA
1. AB=DC
1. Given.
2. BC=AD
2. Given.
3. AC=AC
3. Common.
4. △ABC ≌ △ CDA 4. SSS Congruency Rule
5. ∠ BAC= ∠ ACD ⇒ AB ∥ DC 5. C.P.C.T alt. ∠s are equal formed by lines AB,DC and transversal AC.
6. ∠ ACB=∠ CAD ⇒ BC∥AD 6. c.p.c.t. alt, ∠s are equal formed by lines AD, BC and transversal AC.

Hence, ABCD is a parallelogram.▆

Theorem 4

In a parallelogram, opposite angles are equal.


Given:A parallelogram ABCD.
To prove: ∠A=∠C and ∠B=∠D.

Proof:
Statements
Reasons
1. ∠A+∠B=180° 1.AD∥BC and AB is a transversal, sum of co-interior angles=180°
2.∠B=∠C=180° 2. AB∥DC and BC is a transversal, sum of co-interior angles=180°
3.∠A+∠B=∠B+∠C
⇒ ∠A=∠C
Similarly, ∠B = ∠D.
From 1 and 2


Theorem 5

If each pair of opposite angles of a quadrilateral is equal, then it is a parallelogram.
Given: A quadrilateral ABCD in which ∠A=∠C and ∠B=∠D.
To prove: ABCD is a parallelogram.

Proof:
Statements
Reasons
1. ∠A=∠C 1. Given
2. ∠B=∠D 2. Given
3.∠A+∠B=∠C+∠D 3. Adding 1 and 2
4. ∠A+∠B+∠C+∠D= 360°
4. Sum of angles of a quadrilateral.
5. 2$($∠A+∠B$)$=360°
⇒∠A+∠B= 180°
⇒ BC‖AD
5. Using line 3 i.e. Sum of co-interior angles=180°,
formed by lines BC,AD and transversal AB.
Similarly, AB ‖DC
Hence, ABCD is a parallelogram.



Theorem 6

If a pair of opposite sides of a quadrilateral is equal and parallel, then it is a parallelogram.
Given: A quadrilateral ABCD in which AB‖DC and AB=DC.
To Prove: ABCD is a parallelogram.
Construction:Join AC.

Proof:
Statements
Reasons
In △ABC and △CDA
1. ∠BAC=∠ACD 1. Alt. ∠s, AB‖DC and AC is a transversal.
2. AB=DC
2. Given
3. AC=CA
3. Common
4. △ABC≌△CDA 4. SAS axion of congruency.
5. ∠ACB=∠CAD⇒AD‖BC
Hence, ABCD is a parallelogram.
5. c.p.c.t. Alt, ∠s are equal formed by lines AD, BC and transversal AC.


Theorem 7

If a pair of opposite sides of a quadrilateral is equal and parallel, then it is a parallelogram.
Given: A quadrilateral ABCD in which AB‖DC and AB=DC.
To prove: ABCD is a parallelogram.
Construction: Join AC.

Proof:
Statements
Reasons
In △ABC and △CDA
1. ∠BAC=∠ACD 1. Alt. ∠s, AB‖DC and AC is a transversal.
2. AB=DC
2. Given
3. AC=CA
3. Common
4. △ABC≌△CDA 4. SAS axion of congruency.
5. ∠ACB=∠CAD
 ⇒AD‖BC
Hence, ABCD is a parallelogram.
5. c.p.c.t. Alt, ∠s are equal formed by lines AD, BC and transversal AC.

Theorem 8

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Given: A quadrilateral ABCD whose diagonals AC and BD intersect at O such that OA=OC and OB=OD.
To Prove: ABCD is a parallelogram.

Proof:
Statements
Reasons
In △OAB and △OCD
1. OA=OC
1. Given.
2. OB=OD
2. Given.
3. ∠AOB=∠COD 3. Vert. opp.∠s.
4. △OAB≌△OCD 4. SAS axiom of congruency
5. ∠AOB=∠OCD
⇒∠CAB=∠ACD
⇒ AB∥DC
5. c.p.c.t.
Alt.∠s are equal formed by lines AB,DC and transversal AC.
6. AB=CD
6. c.p.c.t.
7. ABCD is a parallelogram.
7. In quadrilateral ABCD, AB∥DC and AB=DC

Theorem 9

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.
Given. A triangle ABC,E and F are the mid -points of sides AB and AC respectively.
To prove: EF ∥BC and EF=½ BC.
Construction: Through C, draw a line parallel to BA to meet EF produced at D.

Proof:
Statements
Reasons
In △AEF and △CDF

1. AF=CF
1.F is mid-point of AC (given)
2. ∠AFE =∠CFD
2. Vert.opp.∠s

3. ∠EAF= ∠DCF
3. Alt. ∠s,BA∥CD, by the construction
and AC is a transversal.
4. △AEF≌△CDF 4. ASA rule of congruency.
5. EF=FD and AE=CD
5. c.p.c.t.
6. AE=BE
6. E is the mid-point of AB $($given$)$
7. BE=CD
7. From 5 and 6
8. EBCD is a parallelogram.
8. BA∥CD $($const.$)$ and BE=CD $($from step 7$)$
9. EF∥BC and ED=BC
9. Since EBCD is a parallelogram.
10. EF=½ ED
10. Since EF=FD, from 5
11. EF=½BC
11. Since ED=BC, from 9
Hence, EF∥BC and EF= ½ BC.


Theorem 10

Converse of mid-point theorem
The line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Given:A triangle ABC,E is mid-point of AB. Line l drawn through E and parallel to BC meeting AC at F.
To prove. AF=FC
Construction: Through C, draw a line m parallel BA to meet line l at D.

Proof:
Statements
Reasons
1. EBCD is a parallelogram
1. EF∥BC (given), BA∥CD (const.).
2. BE=CD
2. Opp.sides of a parallelogram are equal.
3. EA=BE
3. E is mid-point of AB(given).
4. EA=CD
4. From 2 and 3.
In △AEF and △CDF
5. ∠EAF=∠DCF
5. Alt. ∠s, CD ∥BA and AC is a transversal.
6. ∠EFA=∠DFC 6. Vert.opp.∠s.
7. EA=CD
7. From 4
8. △AEF≌△CDF 8. AAS rule of ccongruency
9. AF=FC
9. c.p.c.t.
Hence, F is mid-point of AC.