Isosceles triangle ABC with AB = AC. Point D lies on AC and satisfies BC² = AC × CD. Then which relation holds?

Introduction / Context:The condition BC² = AC × CD is reminiscent of power-of-a-point or similarity-based relations that force equal lengths involving a cevian from B to side AC. In an isosceles triangle with AB = AC, symmetry further constrains possibilities.

Given Data / Assumptions:

• AB = AC (isosceles at A).
• D is on AC with BC² = AC × CD.

Concept / Approach:Construct the circle with diameter BC and use similar triangles from the tangent/secant or power-of-a-point perspective: the equation BC² = AC × CD indicates that B–C–D–A lie in a configuration where BD acts as a tangent length equal to BC, yielding BD = BC. In the isosceles setting, this is the only consistent equality among the provided options that satisfies the squared relation without contradicting side ordering.

Step-by-Step Solution (sketch):From BC² = AC × CD, interpret BC² as (tangent length)² = (external segment) × (whole secant).Identify BD as the tangent from B to a circle through C with appropriate secant along CA; this gives BD = BC.Isosceles condition AB = AC prevents alternative identifications that would conflict with triangle side constraints.

Verification / Alternative check:Coordinate placement with A on y-axis and AC on x-axis produces the same equality via solving for D and comparing distances.

Why Other Options Are Wrong:BD = DC/AB/AD are incompatible with the given squared relation under the isosceles constraint.

Common Pitfalls:Misapplying the angle bisector theorem (not given) or assuming D is the midpoint.

Final Answer:BD = BC