Write a proof.
1. Prove: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.
2. Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle.
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Proof 1: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.
To prove this, we need to show that if the diagonals of a parallelogram are perpendicular, then all four sides of the parallelogram are congruent.
Here is the proof:
1. Let ABCD be a parallelogram with diagonals AC and BD.
2. Given that AC and BD are perpendicular.
3. Let E be the point of intersection of the diagonals.
4. Since opposite sides of a parallelogram are parallel, we have AB || CD and BC || AD.
5. Because AC and BD are diagonals, they divide the parallelogram into four triangles: AEC, BED, BCD, and ACD.
6. Since opposite sides of a parallelogram are congruent, we have AE = EC and BE = ED.
7. By the Pythagorean theorem, in the right triangles, AEC and BED, we have AE^2 + EC^2 = AC^2 and BE^2 + ED^2 = BD^2.
8. Since AC and BD are perpendicular, we have AC^2 + BD^2 = AD^2 (by the Pythagorean theorem in the right triangle ACD).
9. Substituting AE^2 + EC^2 for AC^2 and BE^2 + ED^2 for BD^2 in step 7, we have AE^2 + EC^2 + BE^2 + ED^2 = AD^2.
10. By the Pythagorean theorem, we have AE^2 + BE^2 = AB^2 and EC^2 + ED^2 = CD^2.
11. Substituting AB^2 for AE^2 + BE^2 and CD^2 for EC^2 + ED^2 in step 9, we have AB^2 + CD^2 = AD^2.
12. Since AB^2 + CD^2 = AD^2 and AB || CD (opposite sides of a parallelogram are parallel), ABDC is a rectangle (by the converse of the Pythagorean theorem).
13. A rectangle with congruent diagonals is a rhombus.
14. Therefore, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.
Proof 2: If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.
To prove this, we need to show that if the diagonals of a parallelogram are congruent, then the opposite angles of the parallelogram are right angles.
Here is the proof:
1. Let ABCD be a parallelogram with diagonals AC and BD.
2. Given that AC and BD are congruent, AC ≅ BD.
3. Let E be the point of intersection of the diagonals.
4. Since opposite sides of a parallelogram are parallel, we have AB || CD and BC || AD.
5. Because AC and BD are diagonals, they divide the parallelogram into four triangles: AEC, BED, BCD, and ACD.
6. Since opposite sides of a parallelogram are congruent, we have AE = EC and BE = ED.
7. By the Side-Angle-Side (SAS) congruence criterion, we have AEC ≅ BED.
8. Therefore, ∠AEC ≅ ∠BED and ∠BEC ≅ ∠AED.
9. Since ∠AEC and ∠BED are corresponding angles of parallel lines, they are congruent (by the Alternate Interior Angles theorem).
10. Therefore, ∠AEC ≅ ∠BED = 90°, as they are a pair of right angles.
11. Since ∠AEC and ∠BED are right angles, opposite angles of the parallelogram ABCD are right angles.
12. A parallelogram with all right angles is a rectangle.
13. Therefore, if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.
Proof for statement 1:
Given: If the diagonals of a parallelogram are perpendicular.
To prove: The parallelogram is a rhombus.
Proof:
Let ABCD be a parallelogram with diagonals AC and BD.
Step 1: Assume the diagonals AC and BD are perpendicular.
Step 2: Draw AC and BD intersecting at O.
Step 3: In triangle ABC, OA is a diagonal and AB is a side of the parallelogram.
Step 4: Similarly, in triangle BCD, OB is a diagonal and BC is a side of the parallelogram.
Step 5: As diagonals of a parallelogram bisect each other, AO = CO and BO = DO.
Step 6: Since AC and BD are perpendicular, triangle AOC and triangle BOD are right-angled.
Step 7: In triangle AOC, AO = CO and OC = OC (common side).
Step 8: By the hypotenuse-leg congruence theorem, triangle AOC is congruent to triangle COA.
Step 9: Similarly, in triangle DOB, triangle BOD is congruent to triangle BDO.
Step 10: From step 8 and step 9, AO = CO = DO = BO.
Step 11: By definition, a rhombus has all sides congruent.
Step 12: From step 10, all sides of the parallelogram ABCD are congruent.
Step 13: Hence, the parallelogram ABCD with perpendicular diagonals is a rhombus.
Proof for statement 2:
Given: If the diagonals of a parallelogram are congruent.
To prove: The parallelogram is a rectangle.
Proof:
Let ABCD be a parallelogram with diagonals AC and BD.
Step 1: Assume the diagonals AC and BD are congruent.
Step 2: Draw AC and BD intersecting at O.
Step 3: In triangle ABC, OA is a diagonal and AB is a side of the parallelogram.
Step 4: Similarly, in triangle BCD, OB is a diagonal and BC is a side of the parallelogram.
Step 5: As diagonals of a parallelogram bisect each other, AO = CO and BO = DO.
Step 6: Since AC and BD are congruent, triangles AOC and BOD are congruent by side-side-side (SSS) congruence.
Step 7: By the definition of congruence, angles OAC and OCA are congruent, and angles OBD and ODB are congruent.
Step 8: Since the opposite angles in a parallelogram are congruent, angles ABC and ADC are congruent.
Step 9: Since opposite sides of parallelograms are parallel, angles ADC and DCB are congruent by alternate interior angles.
Step 10: From step 8 and step 9, angles ABC and DCB are congruent.
Step 11: By definition, a rectangle has four right angles.
Step 12: From step 10, angles ABC and DCB are right angles.
Step 13: Hence, the parallelogram ABCD with congruent diagonals is a rectangle.