We are assuming you know the terms median, angle bisector and altitude but still, just to be sure, we will start our discussion today by defining them:

Median – A line segment joining a vertex of a triangle with the mid-point of the opposite side.

Angle Bisector – A line segment joining a vertex of a triangle with the opposite side such that the angle at the vertex is split into two equal parts.

Altitude – A line segment joining a vertex of a triangle with the opposite side such that the segment is perpendicular to the opposite side.

Usually, medians, angle bisectors and altitudes drawn from the same vertex of a triangle are different line segments. But in special triangles such as isosceles and equilateral, they can overlap. We will now give you some properties which can be very useful.

I.

In an **isosceles triangle** (where base is the side which is not equal to any other side):

- the altitude drawn to the base is the median and the angle bisector;

- the median drawn to the base is the altitude and the angle bisector;

- the bisector of the angle opposite to the base is the altitude and the median.

II.

The reverse is also true. Consider a triangle ABC:

- If angle bisector of vertex A is also the median, the triangle is isosceles such that AB = AC and BC is the base. Hence this angle bisector is also the altitude.

- If altitude drawn from vertex A is also the median, the triangle is isosceles such that AB = AC and BC is the base. Hence this altitude is also the angle bisector.

- If median drawn from vertex A is also the angle bisector, the triangle is isosceles such that AB = AC and BC is the base. Hence this median is also the altitude.

and so on…

III.

In an **equilateral triangle**, each altitude, median and angle bisector drawn from the same vertex, overlap.

Try to prove all these properties on your own. That way, you will not forget them.

A few things this implies:

- Should an angle bisector in a triangle which is also a median be perpendicular to the opposite side? Yes.

- Can we have an angle bisector which is also a median which is not perpendicular? No. Angle bisector which is also a median implies isosceles triangle which implies it is also the altitude.

- Can we have a median from vertex A which is perpendicular to BC but does not bisect the angle A? No. A median which is an altitude implies the triangle is isosceles which implies it is also the angle bisector.

and so on…

Let’s take a quick question on these concepts:

**Question**: What is ∠A in triangle ABC?

Statement 1: The bisector of ∠A is a median in triangle ABC.

Statement 2: The altitude of B to AC is a median in triangle ABC.

**Solution**: We are given a triangle ABC but we don’t know what kind of a triangle it is.

Jump on to the statements directly.

Statement 1: The bisector of ∠A is a median in triangle ABC.

The angle bisector is also a median. This means triangle ABC must be an isosceles triangle such that AB = AC. But we have no idea about the measure of angle A. This statement alone is not sufficient.

Statement 2: The altitude of ∠B to AC is a median in triangle ABC.

The altitude is also a median. This means triangle ABC must be an isosceles triangle such that AB = BC (Note that the altitude is drawn from vertex B here). But we have no idea about the measure of angle A. This statement alone is not sufficient.

Using both statements together, we see that AB = AC = BC. So the triangle is equilateral! So angle A must be 60 degrees. Sufficient!

Answer (C)

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*