Triangle Median and Angle Bisector: Proving Isosceles Properties

To determine if triangle $\triangle ABC$ is isosceles given that $AD$ is the median and it crosses the predominant angle at vertex $A$, consider the following:

• The median $AD$ divides the opposite side $BC$ into two equal parts, $BD = DC$.
• If $AD$ bisects the predominant angle, it implies that $\triangle ABC$ is symmetric about line $AD$.
• In a triangle, if a median also bisects the angle from which it is drawn, the triangle is isosceles.
• The predominant angle typically refers to either the largest angle or an angle with notable symmetry. If $AD$ bisects it, each half must contribute equally to the full angle, indicating symmetry.

Therefore, since the median $AD$ bisects the predominant angle, $\angle BAD = \angle CAD$, leading to the symmetry required for isosceles triangle properties in $\triangle ABC$.

Yes, triangle $ABC$ is indeed isosceles.