Saved Bookmarks
| 1. |
Prove that a parallelogram is a rhombus |
| Answer» \t\tIf all sides of a quadrilateral are congruent, then it’s a rhombus (reverse of the definition).\t\t\tIf the diagonals of a quadrilateral bisect all the angles, then it’s a rhombus (converse of a property).\t\t\tIf the diagonals of a quadrilateral are perpendicular bisectors of each other, then it’s a rhombus (converse of a property).\tTip:\xa0To visualize this one, take two pens or pencils of different lengths and make them cross each other at right angles and at their midpoints. Their four ends must form a diamond shape — a rhombus.\t\t\tIf two consecutive sides of a parallelogram are congruent, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).\t\t\tIf either diagonal of a parallelogram bisects two angles, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).\t\t\tIf the diagonals of a parallelogram are perpendicular, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).\tHere’s a rhombus proof for you. Try to come up with a game plan before reading the two-column proof.Statement 1:Reason for statement 1:\xa0Given.Statement 2:Reason for statement 2:\xa0Opposite sides of a rectangle are congruent.Statement 3:Reason for statement 3:\xa0Given.Statement 4:Reason for statement 4:\xa0Like Divisions Theorem.Statement 5:Reason for statement 5:\xa0All angles of a rectangle are right angles.Statement 6:Reason for statement 6:\xa0All right angles are congruent.Statement 7:Reason for statement 7:\xa0Given.Statement 8:Reason for statement 8:\xa0A midpoint divides a segment into two congruent segments.Statement 9:Reason for statement 9:\xa0SAS, or Side-Angle-Side (4, 6, 8)Statement 10:Reason for statement 10:\xa0CPCTC (Corresponding Parts of Congruent Triangles are Congruent).Statement 11:Reason for statement 11:\xa0Given.Statement 12:Reason for statement 12:\xa0If a triangle is isosceles, then its two legs are congruent.Statement 13:Reason for statement 13:\xa0Transitivity (10 and 12).Statement 14:Reason for statement 14:\xa0If a quadrilateral has four congruent sides, then it’s a rhombus. | |