Question : ABCD is a quadrilateral in which AD = BC. What is \(\angle \text{ABC} + \angle \text{ADC}\) equal to ?
Statement-I : AB is parallel to DC
Statement-II : AD is not parallel to BC
Which one of the following is correct in respect of the above Question and the Statements ?
The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone
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.
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