in the diagram of the rhombus ABCD, mC=110. Use the CPCTC Theorem to prove that the diagonals BD and AC are perpendicular by finding the m<D in DEC.

Question

125
35
70
55

Answer 1

To utilize the CPCTC (Corresponding Parts of Congruent Triangles are Congruent) theorem to prove that the diagonals BD and AC of the rhombus ABCD are perpendicular, we can follow these steps:

1. Properties of a Rhombus: Recall that in a rhombus, the diagonals bisect each other at right angles and also bisect the angles of the rhombus.

2. Given: You are provided that \( m\angle C = 110^\circ \).

3. Angle Calculation: Since the diagonals bisect the angles, we can find \( m\angle D \). \[ m\angle D = m\angle C = 110^\circ \] Therefore, since the diagonals bisect \( m\angle C \) and \( m\angle D \): \[ m\angle A = m\angle B = \frac{180^\circ - 110^\circ}{2} = \frac{70^\circ}{2} = 35^\circ \] Also, since the diagonals bisect the angles of the rhombus: \[ m\angle A = m\angle D \]

4. Sum of Angles in Triangle DEC: In triangle DEC, we have:

   • \( m\angle C = 110^\circ \)
   • Each diagonal divides the angles, and since we know \( m\angle D \), we can find \( m\angle E \) and \( m\angle DEC \).

5. Using CPCTC: By showing that triangles DEC and ACB are congruent (or by any other means that demonstrate that the triangles share a side and have equal angles), you can use CPCTC to confirm that the corresponding angles formed by the diagonals are equal.

6. Conclusion on Perpendicular Diagonals: Since you find that the sum of angles in triangle DEC implies each corresponding angle between the diagonals sum to \( 90^\circ \), you thus arrive at the conclusion that diagonals BD and AC are indeed perpendicular.

Final Result

If you assume \( m<D = m\angle DEC \) corresponds to \( 90^\circ \) such that triangles split from the diagonals give right triangles, then: \[ \text{Therefore, } BD \perp AC \] Thus, the diagonals of rhombus ABCD are perpendicular.

Answer 2

it's a multiple answer question

Answer 3

I apologize for the misunderstanding. Let's try to find the correct option for \( m\angle D \) in triangle DEC, considering that \( m\angle C = 110^\circ \) and the properties of a rhombus.

Properties of the Rhombus

1. In a rhombus, opposite angles are equal, so:

   • \( m\angle A = m\angle C \)
   • \( m\angle B = m\angle D \)

2. The sum of all angles in any quadrilateral is \( 360^\circ \). Thus, for rhombus ABCD: \[ m\angle A + m\angle B + m\angle C + m\angle D = 360^\circ \]

Step 1: Find \( m\angle D \)

Knowing \( m\angle C = 110^\circ \), we can find \( m\angle A + m\angle B \): \[ m\angle D + 110^\circ + m\angle B + m\angle A = 360^\circ \] Since \( m\angle A = m\angle C = 110^\circ \) and \( m\angle B = m\angle D \), we have: \[ 110^\circ + m\angle D + 110^\circ + m\angle D = 360^\circ \] Which simplifies to: \[ 220^\circ + 2m\angle D = 360^\circ \] \[ 2m\angle D = 140^\circ \] \[ m\angle D = 70^\circ \]

Conclusion

Thus, \( m\angle D = 70^\circ \). Now reviewing your multiple-choice options:

• 125
• 35
• 70
• 55

The correct choice for \( m\angle D \) is \( \textbf{70} \).

Proof of Perpendicular Diagonals

Using the computed angles: Since we have established that \( m\angle D = 70^\circ \) and \( m\angle C = 110^\circ \), and taking into account the properties of the rhombus where the diagonals bisect the angles, you could demonstrate that triangles formed with the diagonals are right triangles, thereby showing that diagonals \( AC \) and \( BD \) are perpendicular.

If you need further clarification or more steps, feel free to ask!