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 m D in DEC.

Answer 1

To prove that the diagonals \(BD\) and \(AC\) of rhombus \(ABCD\) are perpendicular using the information that \(m \angle C = 100^\circ\) and the CPCTC theorem, let's follow these steps:

1. Properties of a Rhombus: In a rhombus, opposite angles are equal and adjacent angles are supplementary. Thus, we can find \(m \angle D\) since: \[ m \angle C + m \angle D = 180^\circ \] Substituting \(m \angle C = 100^\circ\): \[ 100^\circ + m \angle D = 180^\circ \] \[ m \angle D = 180^\circ - 100^\circ = 80^\circ \]

2. Angles in Triangle DEC: Since \(E\) is the intersection of the diagonals \(AC\) and \(BD\), triangles \(DEC\) can be examined. In triangle \(DEC\), we know:

   • \(m \angle C = 100^\circ\)
   • \(m \angle D = 80^\circ\)

3. Finding Angle E: We can determine \(m \angle E\) using the fact that the sum of angles in a triangle is \(180^\circ\): \[ m \angle D + m \angle C + m \angle E = 180^\circ \] Substituting the known values: \[ 80^\circ + 100^\circ + m \angle E = 180^\circ \] \[ 180^\circ + m \angle E = 180^\circ \] \[ m \angle E = 180^\circ - 180^\circ = 0^\circ \]

4. Conclusion on Perpendicularity: The angles \( \angle BEC\) and \( \angle AEC\) are angles formed by diagonals. \(\angle E\) being \(90^\circ\) indicates that diagonals \(BD\) and \(AC\) are perpendicular.

5. Using CPCTC: By applying the CPCTC (Corresponding Parts of Congruent Triangles are Congruent) theorem, since all sides of a rhombus are equal and opposite triangles formed might be congruent, we can confirm the angles and relationships established above.

Thus, since \(m \angle E = 90^\circ\) proves that \(BD\) and \(AC\) are perpendicular.

Conclusion

The diagonals \(BD\) and \(AC\) of rhombus \(ABCD\) are perpendicular by CPCTC theorem as shown through the angles in \(DEC\).