Parts of a Triangle: Median in an isosceles triangle - properties

In an isosceles triangle, the median to the base is also the height to the base.

That is, side AD forms a 90° angle with side BC.

That is, two right triangles are created.

Therefore, angle ADC is equal to 90 degrees.

To solve this problem, since $AD$ is a median of triangle $ABC$, the median divides the opposite side $BC$ into two equal segments.

Given $BD = 4$, this means that $DC$ must also be equal to 4.

Therefore, the length of $DC$ is $4$.

To solve this problem, we need to examine the properties of an isosceles triangle where the median is drawn from the vertex to the base.

Given $\triangle ABC$ is isosceles with $AB = AC$, and $AD$ is the median to the base $BC$, it follows certain properties specific to isosceles triangles:

• In an isosceles triangle, the median from the vertex angle is also the altitude and the angle bisector for the angle at the vertex.
• Since $AD$ is the median, $BD = DC$.
• This particular property of isosceles triangles means that $AD$ not only divides the base $BC$ equally but also divides the vertex angle $\angle BAC$ equally, thus forming the angle bisector.

Therefore, the angle $\angle BAD$ is equal to $\angle DAC$ because $AD$ serves as the angle bisector of $\angle BAC$ in the isosceles triangle $\triangle ABC$.

Thus, it is true that $\angle BAD = \angle DAC$ in this triangle.

Yes.

To determine if triangle ABC is isosceles given that AD is the median, we must consider the following properties:

• AD divides the opposite side BC into two equal segments, such that $BD = DC$.
• In general geometry, the fact that AD is a median does not alone imply that triangle ABC is isosceles.

Consider specific properties of isosceles triangles. In an isosceles triangle, a median from the apex (or vertex angle) is also an altitude and an angle bisector. However, these conditions arise under unique circumstances where other equal sides or angles are given or can be proven, not merely from the presence of a median.

Since no additional information indicates that sides AB and AC are equal, or that angles at vertices B and C are equal, we cannot conclude that triangle ABC is isosceles based solely on AD being a median.

Therefore, the correct answer is No.

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.