Question 1169820: Give a formal proof of the following theorem
1. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. (Construct a diagonal).
2. If both pairs of opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram. (Construct a diagonal).
3. The diagonals of a rhombus, (i) bisect each other at right angles, (ii) bisect the angles of the rhombus.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely, let's provide formal proofs for these theorems.
**1. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.**
**Given:** Quadrilateral ABCD with AB || CD and AB = CD.
**To Prove:** ABCD is a parallelogram.
**Proof:**
1. **Construct Diagonal AC:** Draw diagonal AC.
2. **Alternate Interior Angles:** Since AB || CD, ∠BAC = ∠DCA (alternate interior angles are congruent).
3. **Congruent Triangles:** Consider triangles ABC and CDA.
* AB = CD (given)
* AC = AC (common side)
* ∠BAC = ∠DCA (proven in step 2)
* Therefore, ΔABC ≅ ΔCDA (SAS congruence).
4. **Corresponding Parts:** Since ΔABC ≅ ΔCDA, ∠BCA = ∠DAC (CPCTC - Corresponding Parts of Congruent Triangles are Congruent).
5. **Parallel Lines:** Since ∠BCA = ∠DAC, AD || BC (alternate interior angles are congruent, so lines are parallel).
6. **Parallelogram:** Since AB || CD (given) and AD || BC (proven in step 5), ABCD is a parallelogram (definition of a parallelogram).
**2. If both pairs of opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram.**
**Given:** Quadrilateral ABCD with AB = CD and AD = BC.
**To Prove:** ABCD is a parallelogram.
**Proof:**
1. **Construct Diagonal AC:** Draw diagonal AC.
2. **Congruent Triangles:** Consider triangles ABC and CDA.
* AB = CD (given)
* BC = AD (given)
* AC = AC (common side)
* Therefore, ΔABC ≅ ΔCDA (SSS congruence).
3. **Corresponding Parts:** Since ΔABC ≅ ΔCDA, ∠BAC = ∠DCA and ∠BCA = ∠DAC (CPCTC).
4. **Parallel Lines:**
* Since ∠BAC = ∠DCA, AB || CD (alternate interior angles are congruent).
* Since ∠BCA = ∠DAC, AD || BC (alternate interior angles are congruent).
5. **Parallelogram:** Since AB || CD and AD || BC, ABCD is a parallelogram (definition of a parallelogram).
**3. The diagonals of a rhombus, (i) bisect each other at right angles, (ii) bisect the angles of the rhombus.**
**Given:** Rhombus ABCD with diagonals AC and BD intersecting at point E.
**To Prove:**
* (i) AC ⊥ BD (diagonals bisect each other at right angles)
* (ii) AC bisects ∠BAD and ∠BCD, and BD bisects ∠ABC and ∠ADC.
**Proof:**
**(i) Diagonals bisect each other at right angles:**
1. **Rhombus Properties:** In a rhombus, all sides are equal: AB = BC = CD = DA.
2. **Congruent Triangles:** Consider triangles ABE and CBE.
* AB = BC (sides of a rhombus)
* BE = BE (common side)
* AE = CE (Diagonals of a parallelogram bisect eachother, and a rhombus is a parallelogram)
* Therefore, ΔABE ≅ ΔCBE (SSS congruence).
3. **Corresponding Parts:** Since ΔABE ≅ ΔCBE, ∠AEB = ∠CEB (CPCTC).
4. **Linear Pair:** ∠AEB and ∠CEB are a linear pair, so ∠AEB + ∠CEB = 180°.
5. **Right Angles:** Since ∠AEB = ∠CEB and ∠AEB + ∠CEB = 180°, ∠AEB = ∠CEB = 90°. Therefore, AC ⊥ BD.
6. **Bisected Segments:** Since ABCD is a parallelogram, AE = CE and BE = DE. Thus the diagonals bisect eachother.
**(ii) Diagonals bisect the angles of the rhombus:**
1. **Congruent Triangles:** Consider triangles ABE and ADE.
* AB = AD (sides of a rhombus)
* AE = AE (common side)
* BE = DE (diagonals of a parallelogram bisect each other)
* Therefore, ΔABE ≅ ΔADE (SSS congruence).
2. **Corresponding Parts:** Since ΔABE ≅ ΔADE, ∠BAE = ∠DAE (CPCTC). Thus AC bisects angle BAD.
3. **Congruent Triangles:** Consider triangles CBE and CDE.
* BC = CD (sides of a rhombus)
* CE = CE (common side)
* BE = DE (diagonals of a parallelogram bisect each other)
* Therefore, ΔCBE ≅ ΔCDE (SSS congruence).
4. **Corresponding Parts:** Since ΔCBE ≅ ΔCDE, ∠BCE = ∠DCE (CPCTC). Thus AC bisects angle BCD.
5. **Congruent Triangles:** Consider triangles ABE and CBE.
* AB = BC (sides of a rhombus)
* BE = BE (common side)
* AE = CE (diagonals of a parallelogram bisect each other)
* Therefore, ΔABE ≅ ΔCBE (SSS congruence).
6. **Corresponding Parts:** Since ΔABE ≅ ΔCBE, ∠ABE = ∠CBE (CPCTC). Thus BD bisects angle ABC.
7. **Congruent Triangles:** Consider triangles ADE and CDE.
* AD = CD (sides of a rhombus)
* DE = DE (common side)
* AE = CE (diagonals of a parallelogram bisect each other)
* Therefore, ΔADE ≅ ΔCDE (SSS congruence).
8. **Corresponding Parts:** Since ΔADE ≅ ΔCDE, ∠ADE = ∠CDE (CPCTC). Thus BD bisects angle ADC.
Therefore, the diagonals of a rhombus bisect each other at right angles and bisect the angles of the rhombus.
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