Proving quadrilaterals
PLEASE PLEASE PLEASE HELP
*the diagonals of a square bisect the vertex angle
* the diagonal of a rhombus are perpendicular
*each diagonal of a rhombus bisects the opposite angles of the rhombus
help me prove please how ? tell me
1,3. consider that the angles are alternate interior angles formed with the parallel sides.
a simple web search will provide proofs of all of these assertions.
Sure, I can help you with that! To prove the given statements about quadrilaterals, we'll use some basic properties and definitions of these shapes. Let's go through each proof step by step:
1. Diagonals of a Square Bisect the Vertex Angle:
To prove this, we can start by drawing a square and labeling its vertices as A, B, C, and D. The diagonals of the square will be the line segments that connect opposite vertices (e.g., AC and BD).
To show that the diagonals bisect the vertex angle, we need to prove that the angle at each vertex of the square is divided into two equal angles by the diagonal. We can use the fact that the opposite sides of a square are parallel and congruent.
Proof:
- Consider triangle ABC. Since AB = BC (sides of a square are congruent), and AC is the diagonal, we have two congruent triangles - ABC and CBA, by the Side-Side-Side (SSS) criterion for triangle congruence.
- Therefore, the angles at vertices B and C are congruent, making each angle divided in half by the diagonal.
2. Diagonals of a Rhombus are Perpendicular:
Let's take a rhombus ABCD and label its vertices. The diagonals of a rhombus are line segments that connect opposite vertices (e.g., AC and BD).
To prove that the diagonals of a rhombus are perpendicular, we will use the properties of a rhombus and the concept of slope.
Proof:
- A rhombus is a parallelogram, so its opposite sides are parallel and congruent.
- Consider triangles ABC and CDA. Since AB = BC = CD = DA (sides of a rhombus are congruent), these triangles are isosceles, and their base angles at B and D are congruent.
- By the Isosceles Triangle Congruence Theorem, we have angle ABC = angle CDA.
- Now, let's consider the slopes of the lines AC and BD formed by the diagonals. Remember that two lines are perpendicular if their slopes are negative reciprocals of each other.
- The slope of AC is (yC - yA) / (xC - xA), and the slope of BD is (yD - yB) / (xD - xB).
- By substituting the coordinates of A, B, C, and D, we can calculate the slopes of AC and BD.
- If the slopes of AC and BD are negative reciprocals (e.g., m1 * m2 = -1), then the diagonals of the rhombus are perpendicular.
3. Each Diagonal of a Rhombus Bisects the Opposite Angles:
For this proof, we again consider a rhombus ABCD and label its vertices.
Proof:
- Similar to the previous proofs, we know that a rhombus has opposite sides that are parallel and congruent.
- Let's consider triangles ABC and CDA. Since AB = BC = CD = DA, these triangles are isosceles, and their base angles at B and D are congruent.
- Now, let's draw the diagonal AC. This diagonal divides the rhombus into two congruent triangles, ABC and CDA.
- Since these triangles share the side AC and have congruent angle B and angle D, we can conclude that the diagonal AC bisects the opposite angles of the rhombus.
Remember, these proofs are just one way to demonstrate the given statements. There might be alternative approaches or proofs using different properties or theorems.