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

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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.

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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

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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.

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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.

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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.

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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

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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.

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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.

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