If in a triangle the bisector of one angle also bisects the opposite side, the triangle must be of which type?

Introduction / Context:
The Angle Bisector Theorem states that an internal angle bisector divides the opposite side into segments proportional to the adjacent sides. If it bisects the opposite side exactly (creates equal segments), that imposes a specific equality on the adjacent sides.

Given Data / Assumptions:
Let △ABC have angle at A bisected by AD with D on BC, and BD = DC (opposite side bisected).

Concept / Approach:
By the Angle Bisector Theorem, BD/DC = AB/AC. If BD = DC, then BD/DC = 1, hence AB/AC = 1 and AB = AC. That means the triangle is isosceles with AB = AC (vertex at A).

Step-by-Step Solution:

Verification / Alternative check:
Construct an isosceles triangle AB = AC; the angle bisector from A is also the median to BC, confirming the property.

Why Other Options Are Wrong:
Scalene forbids any equal sides; right triangle is about a 90° angle (not implied here); “circle” is not a triangle type.

Common Pitfalls:
Confusing angle bisector with perpendicular bisector; assuming any median is an angle bisector (true only in isosceles).

Final Answer:
Isosceles