To prove that the diagonal of rhombus ABCD bisects the opposite angles, we will consider triangle BDC.
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Identify Known Angles: We are given that \( m\angle BDC = 16^\circ \). Since diagonals of a rhombus bisect each other and because of the properties of a rhombus, we know that \( AB = AD \) and \( BC = CD \).
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Use Properties of a Rhombus: In rhombus ABCD, the diagonals bisect the angles at the vertices. This means: \[ m\angle ABD = m\angle DBC \]
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Consider Triangle BDC: Since triangle BDC has two sides that are equal (i.e., \( BD = DC \)), it is isosceles. This implies that the angles opposite to these sides must be equal: \[ m\angle DBC = m\angle BDC \]
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Using Angle Relationships: The angles \( m\angle DBC \) and \( m\angle BDC \) are both \( 16^\circ \), therefore: \[ m\angle DBC = 16^\circ \]
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Find m ABD: The angle \( m\angle ABD \) is equal to \( m\angle DBC \) because the diagonal bisects the angle: \[ m\angle ABD = m\angle DBC = 16^\circ \]
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Use CPCTC: Since \( m\angle ABD = m\angle DBC \) and both angles formed by diagonal \( DB \) are equal, it follows by the CPCTC (Corresponding Parts of Congruent Triangles are Congruent) Theorem that the diagonal \( DB \) bisects \( \angle DAB \).
Thus, we have shown that the angle \( m\angle ABD \) is \( 16^\circ \), confirming that the diagonal of the rhombus bisects the angles. Therefore, the diagonal indeed bisects opposite angles, completing the proof.