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 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 \(DEC\).