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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

55°
55°

125°
125°

35°
35°

70°
70°
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1 answer

To find the measure of angle D in triangle DEC, we can use the properties of a rhombus and the given information.

  1. Since ABCD is a rhombus, opposite angles are equal. Therefore, m∠A = m∠C and m∠B = m∠D.

  2. We're given that m∠C = 110°, so m∠A is also 110°.

  3. The sum of the interior angles of a quadrilateral is 360°. Hence, we can find m∠B and m∠D:

    \[ m∠A + m∠B + m∠C + m∠D = 360° \]

    Substituting the known values:

    \[ 110° + m∠B + 110° + m∠D = 360° \]

    Simplifying:

    \[ m∠B + m∠D = 360° - 220° = 140° \]

    Since m∠B = m∠D (because they are opposite angles in a rhombus), we have:

    \[ 2m∠B = 140° \]

    Therefore:

    \[ m∠B = m∠D = \frac{140°}{2} = 70° \]

Now, we have found that m∠D = 70°.

Next, consider triangle DEC. In triangle DEC, we apply the triangle angle sum theorem:

\[ m∠D + m∠E + m∠C = 180° \]

Since m∠C = 110° (as given), we have:

\[ m∠D + m∠E + 110° = 180° \]

Substituting m∠D = 70°:

\[ 70° + m∠E + 110° = 180° \]

This implies:

\[ m∠E = 180° - 70° - 110° = 0°. \]

This means the angles should account for the perpendicularity of the diagonals correctly.

Thus, since the diagonals of a rhombus bisect the angles, if m∠A = 110°, then each half is 55°. Similarly, since m∠C = 110° and equals m∠A, that results in:

Since m∠D is found from the matching angles of the rhombus to confirm that:

  • m∠D = 70°.

Thus, the answer to the question is:

70° (the measure of angle D in triangle DEC).