Question

ABCD is a quadrilateral in which AD = BC and \(\angle \text{DAB} = \angle \text{CBA}\). If \(\angle \text{CAB} = 30^\circ\), then the measure of \(\angle \text{AOB}\) is :

• A. \(80^\circ\)
• B. \(100^\circ\)
• C. \(120^\circ\)
• D. \(135^\circ\)

Hint:

1. Focus on \(\triangle \text{DAB}\) and \(\triangle \text{CBA}\). 2. Use the given AD = BC, \(\angle \text{DAB} = \angle \text{CBA}\), and common side AB to prove \(\triangle \text{DAB} \cong \triangle \text{CBA}\) by SAS. 3. Congruence means corresponding angles are equal: \(\angle \text{DBA} = \angle \text{CAB}\). 4. Since \(\angle \text{CAB} = 30^\circ\), then \(\angle \text{DBA} = 30^\circ\). 5. In \(\triangle \text{AOB}\) (where O is the intersection of diagonals), you now have \(\angle \text{OAB} = 30^\circ\) and \(\angle \text{OBA} = 30^\circ\). 6. Find \(\angle \text{AOB}\) using the sum of angles in \(\triangle \text{AOB}\): \(180^\circ - (30^\circ + 30^\circ) = 120^\circ\).

Answer 1

The Correct Option is C

Concept: This problem uses properties of congruent triangles and angles in a triangle. Point O is assumed to be the intersection of the diagonals AC and BD, as typically shown in such diagrams. Step 1: Analyze the given conditions to prove triangle congruence Given for quadrilateral ABCD:
AD = BC
\(\angle \text{DAB} = \angle \text{CBA}\) Consider \(\triangle \text{DAB}\) and \(\triangle \text{CBA}\). (A) AD = BC (Given side) (B) \(\angle \text{DAB} = \angle \text{CBA}\) (Given angle) (C) AB = BA (Common side) By the SAS (Side-Angle-Side) congruence criterion, \(\triangle \text{DAB} \cong \triangle \text{CBA}\). Step 2: Use consequences of congruent triangles Since \(\triangle \text{DAB} \cong \triangle \text{CBA}\), their corresponding parts are equal. Particularly, corresponding angles: \[ \angle \text{DBA} = \angle \text{CAB} \] (Also, diagonals AC = BD). Step 3: Use the given angle \(\angle \text{CAB}\) Given \(\angle \text{CAB} = 30^\circ\). From the congruence in Step 2, we have \(\angle \text{DBA} = \angle \text{CAB}\). Therefore, \(\angle \text{DBA} = 30^\circ\). Step 4: Consider \(\triangle \text{AOB}\) Assuming O is the intersection of the diagonals AC and BD (as implied by the question asking for \(\angle AOB\) and typical diagrams for such problems). In \(\triangle \text{AOB}\):
\(\angle \text{OAB}\) is the same as \(\angle \text{CAB} = 30^\circ\).
\(\angle \text{OBA}\) is the same as \(\angle \text{DBA} = 30^\circ\). The sum of angles in \(\triangle \text{AOB}\) is \(180^\circ\). \[ \angle \text{AOB} + \angle \text{OAB} + \angle \text{OBA} = 180^\circ \] Substitute the known values: \[ \angle \text{AOB} + 30^\circ + 30^\circ = 180^\circ \] \[ \angle \text{AOB} + 60^\circ = 180^\circ \] \[ \angle \text{AOB} = 180^\circ - 60^\circ \] \[ \angle \text{AOB} = 120^\circ \] This matches option (3). (Note: A quadrilateral with AD=BC and \(\angle DAB = \angle CBA\) implies that its diagonals are equal, AC=BD. This type of quadrilateral is often an isosceles trapezoid if the non-parallel sides AD and BC are equal, or it could be a rectangle. The congruence of \(\triangle DAB\) and \(\triangle CBA\) is key.)