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Question

A rhombus labeled upper A upper B upper C upper D has 2 interior, diagonal lines.

In the diagram of the rhombus ABCD, m∠C=110°. Use the CPCTC Theorem to prove that the diagonals BD¯¯¯¯¯¯¯¯ and AC¯¯¯¯¯¯¯¯ are perpendicular by finding the m∠D in △DEC.

(1 point)
Responses

70°
70°

125°
125°

35°
35°

55°
55°
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Answer 1

To find \( m∠D \) in triangle \( DEC \) and demonstrate that the diagonals \( BD \) and \( AC \) are perpendicular using the CPCTC theorem, we can follow these steps:

1. Since the sum of the angles in triangle \( ABC \) must total 180°, we can find \( m∠A \) and \( m∠B \) because \( ABCD \) is a rhombus, which means opposite angles are equal.

   • \( m∠A = m∠C = 110° \)
   • Since the opposite angles of a rhombus are also equal, \( m∠B = m∠D \).

2. The sum of all angles in the rhombus is \( 360° \). For triangle \( ABC \): \[ m∠A + m∠B + m∠C + m∠D = 360° \] Substituting \( m∠A \) and \( m∠C \): \[ 110° + m∠B + 110° + m∠D = 360° \] Simplifying gives: \[ 220° + m∠B + m∠D = 360° \]

3. Therefore, we have: \[ m∠B + m∠D = 360° - 220° = 140° \]

4. Since \( m∠B = m∠D \), we can set up the equation: \[ 2m∠B = 140° \] Which simplifies to: \[ m∠B = 70° \] Thus, \( m∠D = 70° \).

5. Since \( AC \) and \( BD \) are diagonals, we can use the property that in a rhombus, the diagonals bisect each other at right angles. Because \( m∠B + m∠D = 140° \), both angles being equal at 70° implies \( ∠B + ∠D = 180°.\)

6. Therefore, the diagonals \( AC \) and \( BD \) are perpendicular to each other.

The answer is:

70°.