Question

In the figure (not drawn to scale) given below, if \(AD = CD = BC\) and \(\angle BCE = 96^\circ\), how much is the value of \(\angle DBC\)?

• A. \(32^\circ\)
• B. \(84^\circ\)
• C. \(64^\circ\)
• D. Cannot be determined

Hint:

Use the isosceles triangles ACD and DCB, and the exterior angle of a triangle at C.

Answer 1

The Correct Option is C

Step 1: Set up the figure:
A, D, B lie on a straight line, and E lies on the extension of AC beyond C. Let \(\angle CAD = x\).

Step 2: Use triangle ACD:
Since \(AD = CD\), the angles opposite these sides are equal, so \(\angle DCA = \angle DAC = x\). The exterior angle \(\angle CDB\) of triangle ACD equals the sum of the two opposite interior angles, so \(\angle CDB = x + x = 2x\).

Step 3: Use triangle CDB:
Since \(CD = BC\), we get \(\angle CBD = \angle CDB = 2x\).

Step 4: Use the exterior angle at C of triangle ABC:
\(\angle BCE = \angle CAB + \angle CBA = x + 2x = 3x\). Given \(\angle BCE = 96^\circ\), so \(3x = 96^\circ\) and \(x = 32^\circ\).

Step 5: Find the required angle:
\(\angle DBC = 2x = 64^\circ\).

Final Answer:
The correct option is (3): \(64^\circ\).