# 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.

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 ‖DCHence, 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.