Analyzing Quadrilateral Properties for Angle Sum

The question asks for the value of \(\angle \text{ABC} + \angle \text{ADC}\) given that ABCD is a quadrilateral with AD = BC.

Statement I Analysis

Statement I says AB is parallel to DC (AB || DC).

Given AD = BC and AB || DC, the quadrilateral ABCD is either a rectangle or an isosceles trapezoid.

• If ABCD is a rectangle, all angles are \(90^\circ\). Then \(\angle \text{ABC} = 90^\circ\) and \(\angle \text{ADC} = 90^\circ\). The sum is \(90^\circ + 90^\circ = 180^\circ\).
• If ABCD is an isosceles trapezoid with bases AB and DC, then \(\angle \text{DAB} = \angle \text{CBA}\) and \(\angle \text{ADC} = \angle \text{BCD}\). Also, consecutive angles between the parallel bases are supplementary. Thus, \(\angle \text{DAB} + \angle \text{ADC} = 180^\circ\) and \(\angle \text{CBA} + \angle \text{BCD} = 180^\circ\).
• We need \(\angle \text{ABC} + \angle \text{ADC}\). Let \(\angle \text{ABC} = \alpha\). Since \(\angle \text{CBA} + \angle \text{BCD} = 180^\circ\), we have \(\alpha + \angle \text{BCD} = 180^\circ\). Since \(\angle \text{ADC} = \angle \text{BCD}\) in an isosceles trapezoid, \(\angle \text{ADC} = 180^\circ - \alpha\).
• Therefore, \(\angle \text{ABC} + \angle \text{ADC} = \alpha + (180^\circ - \alpha) = 180^\circ\).

Conclusion: Statement I alone is sufficient to determine that \(\angle \text{ABC} + \angle \text{ADC} = 180^\circ\).

Statement II Analysis

Statement II says AD is not parallel to BC.

This condition alone, even with AD = BC, does not uniquely determine the quadrilateral or the required angle sum.

Consider a quadrilateral ABCD with AD = BC = 5. If AD is not parallel to BC, the shape can vary significantly. For example, consider vertices A=(0,3), B=(4,0), C=(0,-3), D=(-4,0). Here AD = BC = 5. AD is not parallel to BC. Calculating the angles: Vector BA = (-4, 3), Vector BC = (-4, -3). \(\cos(\angle \text{ABC}) = \frac{(-4)(-4) + (3)(-3)}{\sqrt{16+9}\sqrt{16+9}} = \frac{16-9}{25} = \frac{7}{25}\). Vector DA = (4, 3), Vector DC = (4, -3). \(\cos(\angle \text{ADC}) = \frac{(4)(4) + (3)(-3)}{\sqrt{16+9}\sqrt{16+9}} = \frac{16-9}{25} = \frac{7}{25}\). So \(\angle \text{ABC} = \angle \text{ADC} = \arccos(7/25)\). The sum is \(2 \arccos(7/25)\), which is not \(180^\circ\).

Conclusion: Statement II alone is not sufficient.

Conclusion on Both Statements

Statement I alone is sufficient to answer the question.

Statement II alone is not sufficient.

Therefore, the question can be answered using Statement I alone, but not using Statement II alone.

This matches Option 1.