In a rhombus, the diagonals bisect each other at right angles. To prove that the diagonals \(BD\) and \(AC\) are perpendicular using the given information, we can analyze the measures of the angles involved.
From the problem, you are given that \(mC = 110^\circ\). Since \(ABCD\) is a rhombus, opposite angles are equal, which means:
- \(mA = mC = 110^\circ\)
- \(mB = mD\)
The sum of the angles in a quadrilateral is \(360^\circ\). Therefore, we can find the measures of angles \(B\) and \(D\) as follows:
\[ mA + mB + mC + mD = 360^\circ \] Substituting the known values:
\[ 110^\circ + mB + 110^\circ + mD = 360^\circ \] \[ 220^\circ + mB + mD = 360^\circ \]
This implies:
\[ mB + mD = 360^\circ - 220^\circ = 140^\circ \]
Since \(mB = mD\) (because opposite angles in a rhombus are equal):
\[ 2mB = 140^\circ \] \[ mB = 70^\circ \] Thus, \(mD = 70^\circ\).
Now, the angles \(BCD\) and \(DAB\) are supplementary because they are consecutive angles in the rhombus:
\[ mB + mD = 70^\circ + 70^\circ = 140^\circ \]
Next, since diagonals bisect angles in a rhombus:
\[ mCDB = \frac{mC}{2} = \frac{110^\circ}{2} = 55^\circ \] \[ mA = mDAC = 55^\circ \]
Now we can say \( \triangle ABC \) and \( \triangle ADC \) are congruent through which we utilize CPCTC (Corresponding Parts of Congruent Triangles are Congruent) with:
- \(mCDA = 55^\circ\)
- \(mBCD = 55^\circ\)
Since:
- \(mCDB + mBCD = 110^\circ\)
- \(mDAB + mCDA = 110^\circ\)
Since diagonals bisect the angles and \( \triangle ABC \) and \( \triangle ADC\) are congruent, \(AC\) and \(BD\) meet at a right angle, confirming that they are perpendicular.
Thus, we have proven that the diagonals \(BD\) and \(AC\) of rhombus \(ABCD\) are perpendicular.
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