Proofs of theorems related to Parallelograms.
Theorem 1
A diagonal of a parallelogram divides it into two congruent triangles.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 
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. 
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. 
Statements 
Reasons 
1. ∠A+∠B=180°  1.AD∥BC and AB is a transversal, sum of cointerior angles=180° 
2.∠B=∠C=180°  2. AB∥DC and BC is a transversal, sum of cointerior angles=180° 
3.∠A+∠B=∠B+∠C ⇒ ∠A=∠C Similarly, ∠B = ∠D. 
From 1 and 2 
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 cointerior angles=180°, formed by lines BC,AD and transversal AB. 
Similarly, AB ‖DC Hence, ABCD is a parallelogram. 
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. 
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. 
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 
Statements 
Reasons 
In △AEF and △CDF 

1. AF=CF 
1.F is midpoint 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 midpoint 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. 
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 midpoint 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 midpoint of AC. 