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 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 |
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. |
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 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. |
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. |