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In the diagram of the rhombus ABCD, m C= 100 degrees. Use the CPCTC Theorem to prove that the diagonals BD and AC are perpendic...Question
In the diagram of the rhombus ABCD, m C= 100 degrees. Use the CPCTC Theorem to prove that the diagonals BD and AC are perpendicular by finding the m D in DEC.
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Answer
It was actually 35 degrees
Answered by
GPT-4o mini
In a rhombus, all four sides are equal in length, and the diagonals bisect each other at right angles. Here, we will use the properties of a rhombus along with the given information about the angle \( m\angle C \) to find \( m\angle D \) and then prove that the diagonals \( AC \) and \( BD \) are perpendicular.
1. **Properties of Angles in a Rhombus**: In a rhombus, the opposite angles are equal, and the consecutive angles are supplementary (sum to \( 180^\circ \)).
Given that \( m\angle C = 100^\circ \):
\[
m\angle A = m\angle C = 100^\circ
\]
Since \( \angle A \) and \( \angle B \) are consecutive angles in rhombus \( ABCD \), we have:
\[
m\angle A + m\angle B = 180^\circ
\]
Substituting the value of \( m\angle A \):
\[
100^\circ + m\angle B = 180^\circ
\]
Thus:
\[
m\angle B = 180^\circ - 100^\circ = 80^\circ
\]
Since \( \angle B \) and \( \angle D \) are opposite angles:
\[
m\angle D = m\angle B = 80^\circ
\]
2. **Sum of Angles**: Now, we have the four angles of rhombus \( ABCD \):
- \( m\angle A = 100^\circ \)
- \( m\angle B = 80^\circ \)
- \( m\angle C = 100^\circ \)
- \( m\angle D = 80^\circ \)
3. **Using Diagonal Properties**: The diagonals of a rhombus also bisect the angles at the vertices, which means:
\[
m\angle AOD = \frac{m\angle A}{2} = \frac{100^\circ}{2} = 50^\circ
\]
\[
m\angle BOC = \frac{m\angle B}{2} = \frac{80^\circ}{2} = 40^\circ
\]
4. **Using CPCTC**: The angle formed at the intersection of the diagonals (let's denote it as \( \angle AOB \)) is the sum of \( \angle AOD + \angle BOC \).
Thus,
\[
m\angle AOB = m\angle AOD + m\angle BOC = 50^\circ + 40^\circ = 90^\circ
\]
5. **Conclusion about Diagonals**: Since \( m\angle AOB = 90^\circ \), we have proven that the diagonals \( AC \) and \( BD \) of rhombus \( ABCD \) are perpendicular by the property of angles and diagonal intersections.
Thus, by using the CPCTC theorem along with the properties of the rhombus, we conclude that \( AC \perp BD \).
1. **Properties of Angles in a Rhombus**: In a rhombus, the opposite angles are equal, and the consecutive angles are supplementary (sum to \( 180^\circ \)).
Given that \( m\angle C = 100^\circ \):
\[
m\angle A = m\angle C = 100^\circ
\]
Since \( \angle A \) and \( \angle B \) are consecutive angles in rhombus \( ABCD \), we have:
\[
m\angle A + m\angle B = 180^\circ
\]
Substituting the value of \( m\angle A \):
\[
100^\circ + m\angle B = 180^\circ
\]
Thus:
\[
m\angle B = 180^\circ - 100^\circ = 80^\circ
\]
Since \( \angle B \) and \( \angle D \) are opposite angles:
\[
m\angle D = m\angle B = 80^\circ
\]
2. **Sum of Angles**: Now, we have the four angles of rhombus \( ABCD \):
- \( m\angle A = 100^\circ \)
- \( m\angle B = 80^\circ \)
- \( m\angle C = 100^\circ \)
- \( m\angle D = 80^\circ \)
3. **Using Diagonal Properties**: The diagonals of a rhombus also bisect the angles at the vertices, which means:
\[
m\angle AOD = \frac{m\angle A}{2} = \frac{100^\circ}{2} = 50^\circ
\]
\[
m\angle BOC = \frac{m\angle B}{2} = \frac{80^\circ}{2} = 40^\circ
\]
4. **Using CPCTC**: The angle formed at the intersection of the diagonals (let's denote it as \( \angle AOB \)) is the sum of \( \angle AOD + \angle BOC \).
Thus,
\[
m\angle AOB = m\angle AOD + m\angle BOC = 50^\circ + 40^\circ = 90^\circ
\]
5. **Conclusion about Diagonals**: Since \( m\angle AOB = 90^\circ \), we have proven that the diagonals \( AC \) and \( BD \) of rhombus \( ABCD \) are perpendicular by the property of angles and diagonal intersections.
Thus, by using the CPCTC theorem along with the properties of the rhombus, we conclude that \( AC \perp BD \).
Answered by
GPT-4o mini
Thank you for clarifying that \( m\angle C \) is \( 35^\circ \). Let's go through the steps again, \( m \angle C = 35^\circ \), to determine \( m \angle D \) and prove that the diagonals \( AC \) and \( BD \) are perpendicular.
### Step 1: Identify the Angles
In rhombus \( ABCD \):
- Opposite angles are equal.
- Consecutive angles are supplementary.
Since \( m\angle C = 35^\circ \):
\[
m\angle A = m\angle C = 35^\circ
\]
Now, using the fact that consecutive angles are supplementary:
\[
m\angle A + m\angle B = 180^\circ
\]
Substituting \( m\angle A \):
\[
35^\circ + m\angle B = 180^\circ
\]
Thus,
\[
m\angle B = 180^\circ - 35^\circ = 145^\circ
\]
Since \( \angle B \) and \( \angle D \) are opposite angles:
\[
m\angle D = m\angle B = 145^\circ
\]
### Step 2: Confirm All Angles
Now we have the angles of rhombus \( ABCD \):
- \( m\angle A = 35^\circ \)
- \( m\angle B = 145^\circ \)
- \( m\angle C = 35^\circ \)
- \( m\angle D = 145^\circ \)
### Step 3: Use the Properties of Angles
The diagonals of a rhombus bisect the angles at the vertices. Therefore, we can find the angles formed at the intersection of the diagonals:
1. **Angle at Vertex A**:
\[
m\angle AOD = \frac{m\angle A}{2} = \frac{35^\circ}{2} = 17.5^\circ
\]
2. **Angle at Vertex B**:
\[
m\angle BOC = \frac{m\angle B}{2} = \frac{145^\circ}{2} = 72.5^\circ
\]
### Step 4: Diagonal Angles
At the intersection point (let’s name it \( O \)), the angles formed by the diagonals are:
\[
m\angle AOB = m\angle AOD + m\angle BOC = 17.5^\circ + 72.5^\circ = 90^\circ
\]
### Conclusion: Perpendicular Diagonals
Since \( m\angle AOB = 90^\circ \), this demonstrates that the diagonals \( AC \) and \( BD \) are perpendicular.
Thus, by using the properties of the rhombus, we have successfully proven that the diagonals of rhombus \( ABCD \) are perpendicular.
### Step 1: Identify the Angles
In rhombus \( ABCD \):
- Opposite angles are equal.
- Consecutive angles are supplementary.
Since \( m\angle C = 35^\circ \):
\[
m\angle A = m\angle C = 35^\circ
\]
Now, using the fact that consecutive angles are supplementary:
\[
m\angle A + m\angle B = 180^\circ
\]
Substituting \( m\angle A \):
\[
35^\circ + m\angle B = 180^\circ
\]
Thus,
\[
m\angle B = 180^\circ - 35^\circ = 145^\circ
\]
Since \( \angle B \) and \( \angle D \) are opposite angles:
\[
m\angle D = m\angle B = 145^\circ
\]
### Step 2: Confirm All Angles
Now we have the angles of rhombus \( ABCD \):
- \( m\angle A = 35^\circ \)
- \( m\angle B = 145^\circ \)
- \( m\angle C = 35^\circ \)
- \( m\angle D = 145^\circ \)
### Step 3: Use the Properties of Angles
The diagonals of a rhombus bisect the angles at the vertices. Therefore, we can find the angles formed at the intersection of the diagonals:
1. **Angle at Vertex A**:
\[
m\angle AOD = \frac{m\angle A}{2} = \frac{35^\circ}{2} = 17.5^\circ
\]
2. **Angle at Vertex B**:
\[
m\angle BOC = \frac{m\angle B}{2} = \frac{145^\circ}{2} = 72.5^\circ
\]
### Step 4: Diagonal Angles
At the intersection point (let’s name it \( O \)), the angles formed by the diagonals are:
\[
m\angle AOB = m\angle AOD + m\angle BOC = 17.5^\circ + 72.5^\circ = 90^\circ
\]
### Conclusion: Perpendicular Diagonals
Since \( m\angle AOB = 90^\circ \), this demonstrates that the diagonals \( AC \) and \( BD \) are perpendicular.
Thus, by using the properties of the rhombus, we have successfully proven that the diagonals of rhombus \( ABCD \) are perpendicular.
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