Angle chase with isosceles condition and external division:\nIn ΔABC where AC = BC and ∠ABC = 50°, extend BC to point D so that BC = CD. Find the value of ∠BAD.

Introduction / Context:
This is a classic Euclidean angle-chasing configuration combining an isosceles triangle (AC = BC) with an external point D chosen so that BC = CD. The task is to find the angle at A between AB and AD.

Given Data / Assumptions:

• AC = BC (isosceles with apex at C).
• ∠ABC = 50° (angle at B).
• BC is produced to D with BC = CD.

Concept / Approach:
One method is coordinate/analytic geometry: set B = (0,0), C = (1,0), so D = (2,0). Choose A so that AC = 1 and ∠ABC = 50°. Alternatively, pure geometry using isosceles properties and external point reflection arguments also leads to a right angle at ∠BAD.

Step-by-Step Solution (analytic sketch):

Verification / Alternative check:
Known result: with AC = BC and external point D such that BC = CD, the locus creates a right angle at A between AB and AD due to reflective symmetry about the perpendicular at the midpoint of BD.

Why Other Options Are Wrong:
80°, 40°, 50°, 70° do not satisfy the orthogonality deduction; only 90° is consistent with both construction and computations.

Common Pitfalls:
Assuming interior angle sums directly determine ∠BAD without constructing D carefully, or overlooking the equality BC = CD which enforces a key symmetry.

Final Answer:
90°