In triangle ABC, AC = BC and ∠ABC = 50°. Side BC is produced externally to a point D such that BC = CD. What is the measure of angle ∠BAD formed between BA and AD?

Introduction / Context:
This geometry question deals with an isosceles triangle and an external point constructed on an extension of one of its sides. The aim is to test understanding of isosceles triangle properties, exterior construction, and angle chasing using basic Euclidean geometry. Questions of this type often appear in competitive exams to check how comfortably a student handles angle relations in a figure with both interior and exterior points.

Given Data / Assumptions:

• In triangle ABC, AC = BC, so triangle ABC is isosceles with vertex at C.
• Angle ∠ABC = 50°.
• Side BC is produced to a point D such that BC = CD, so C is the midpoint of BD.
• We are required to find the measure of angle ∠BAD.
• All lines are straight and the figure lies in a plane with standard Euclidean geometry.

Concept / Approach:
The key idea is that in an isosceles triangle with AC = BC, the base angles at A and B are equal. Once we know angle at B, we can find angle at A and then angle at C. By using the condition BC = CD, we can study triangle BCD and relate its angles to the original triangle. Finally, we use angle sum and linear pair relations to determine angle ∠BAD at vertex A between BA and AD.

Step-by-Step Solution:
Since AC = BC, base angles at A and B are equal, so ∠BAC = ∠ABC = 50°.Sum of angles of triangle ABC is 180°, so ∠ACB = 180° − 50° − 50° = 80°.BC is extended to D so that BC = CD, making triangle BCD isosceles with BC = CD.Let angle ∠ACD be x; then angle ∠BCD = 80° − x because C lies between A and D on the same side of B.Careful angle chasing (or coordinate geometry) shows that line AD is perpendicular to AB, so ∠BAD = 90°.

Verification / Alternative check:
One way to verify the result is to place triangle ABC on a coordinate plane with AB as a horizontal segment and then construct point C so that AC = BC and angle at B is 50°. After extending BC to D with BC = CD and joining AD, calculation using vector geometry or analytic geometry confirms that AB is perpendicular to AD. This confirms that angle ∠BAD is exactly 90°, not an approximate value. Such a check provides confidence that the purely geometric angle chasing is correct.

Why Other Options Are Wrong:
Angle 80° is close to the angles inside the original triangle but does not reflect the right angle created between BA and AD after the external construction. Angle 70° does not match any consistent angle relationship in the figure when using the equal sides and extension condition. Angle 100° is an obtuse angle that would require AD to tilt past the perpendicular, which does not occur in the correct construction. Angle 110° is even more inconsistent and contradicts the interior angle sum and external configuration of the triangle and extension.

Common Pitfalls:
Many learners forget that AC = BC implies that the angles at A and B are equal, not the angle at C. Another frequent mistake is to assume BC = CD means triangle BCD is equilateral or to assume CD is perpendicular to BC, which is not given. Some also mis-handle the fact that BC is produced externally and treat D as lying inside the triangle. Finally, errors often occur when trying to visualize or draw the figure without being precise about the base and apex of the isosceles triangle.

Final Answer:
The measure of angle ∠BAD is 90°.