SSC CGL Maths · Triangles

Medians and Apollonius Theorem

A median joins a corner to the midpoint of the opposite side; the three medians meet at the centroid, two thirds of the way along each, and Apollonius' theorem gives a median's length from the sides.

All Triangles notes

Why this matters

Ten questions in the bank, 9 EASY and 1 MODERATE, four of them from the 2025 shifts. Four use the 2 : 1 split at the centroid, three the equal areas that medians make, and three Apollonius' theorem or the median to a hypotenuse. They are quick marks once the 2 : 1 is read from the right end.

Concept 1 of 3: The centroid divides each median in the ratio 2 : 1

The centroid is the balance point of a triangle cut from card. It sits on every median, twice as far from the corner as from the midpoint of the opposite side.

Definition

A median joins a corner to the midpoint of the opposite side. The three medians meet at one point, the centroid G.
G divides each median 2 : 1 from the corner: on the median AD, $AG = \frac{2}{3}AD$ and $GD = \frac{1}{3}AD$ , so $AD = 3 \times GD$ .
Equilateral triangle of side $a$ : each median is also the height, $\frac{\sqrt{3}}{2}a$ , so the distance from a corner to the centroid is $\frac{2}{3} \times \frac{\sqrt{3}}{2}a = \frac{a}{\sqrt{3}}$ .
Two medians at right angles: if AD and BE meet at G at 90°, triangle AGB is right-angled with legs $AG = \frac{2}{3}AD$ and $BG = \frac{2}{3}BE$ , so Pythagoras gives AB.

Centroid on a median

AG = \frac{2}{3}AD \qquad GD = \frac{1}{3}AD

ADthe median from A to the midpoint D of BC
Gthe centroid

Worked example

In triangle PQR, the median PM is 21 cm long and G is the centroid. Find PG and GM.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 16 Sep 2025, Shift 1 · Q69Easy

Example 1 · Triangles · Medians and Apollonius Theorem

In a triangle ABC, medians AD and BE intersect at G. If the length of median AD is 12 cm, what is the length of the segment AG?

Reading 2 : 1 from the wrong end

The longer part of a median is the one at the corner. GD, at the midpoint end, is one third of the median.

Using whole medians when two medians cross at 90°

The right angle is at the centroid, so the right triangle's legs are the two-thirds parts AG and BG, not the full medians.

Drill 3 more on the centroid divides each median in the ratio 2 : 1

Concept 2 of 3: A median halves the area, and the centroid makes equal parts

Two triangles with the same height have areas in the ratio of their bases. A median gives two equal bases under one height, so it cuts the area exactly in half.

Definition

Same height, compare the bases: triangles with their top corner at A and bases on BC have areas in the ratio of their bases.
So a median AD halves the area: ar(ABD) = ar(ACD) = $\frac{1}{2}$ ar(ABC).
The three medians cut a triangle into six small triangles of equal area, each $\frac{1}{6}$ of the whole.
So GAB, GBC and GCA each have $\frac{1}{3}$ of the area, where G is the centroid.

Areas made by medians

\text{ar}(ABD) = \frac{1}{2}\,\text{ar}(ABC) \qquad \text{ar}(GBC) = \frac{1}{3}\,\text{ar}(ABC)

Worked example

Triangle ABC has an area of 72 cm². AD is a median and G is the centroid. Find the areas of triangles ABD, GBC and GBD.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 3 · Q68Easy

Example 2 · Triangles · Medians and Apollonius Theorem

In triangle ABC, medians AD, BE, and CF intersect at the Centroid G. If the area of triangle ABC is

60 \text{ cm}^2

, what is the area of triangle GBC?

Taking the centroid triangles as quarters

GAB, GBC and GCA are each a third of the triangle. The quarter belongs to the triangle that joins the midpoints of the sides.

Thinking a median quarters the area

A median halves one side and keeps the height, so it halves the area.

Drill 2 more on a median halves the area, and the centroid makes equal parts

Concept 3 of 3: Apollonius theorem: the length of a median

Drop the median AD onto BC and the two sides AB and AC lean out on either side of it. Apollonius' theorem ties their squares to the median and the half-side, the way Pythagoras ties the sides of a right triangle.

Definition

For the median AD to BC: $AB^2 + AC^2 = 2(AD^2 + BD^2)$ , where $BD = \frac{1}{2}BC$ .
Solved for the median: $AD = \frac{1}{2}\sqrt{2AB^2 + 2AC^2 - BC^2}$ .
Median to the hypotenuse: if $\angle A = 90^\circ$ , then $AB^2 + AC^2 = BC^2$ and the theorem gives $AD = \frac{1}{2}BC$ . The midpoint of the hypotenuse is the same distance from all three corners.
Two equal medians come from two equal sides, so a triangle with all three medians equal is equilateral.

Apollonius theorem

AB^2 + AC^2 = 2(AD^2 + BD^2) \qquad AD = \frac{1}{2}\sqrt{2AB^2 + 2AC^2 - BC^2}

Dthe midpoint of BC, so BD is half of BC

Worked example

In triangle ABC, AB = 7 cm, AC = 9 cm and BC = 8 cm. Find the median AD.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 19 Sep 2024, Shift 3 · Q62Easy

Example 3 · Triangles · Medians and Apollonius Theorem

\Delta PQR

, if PT is the median, then which of the following is correct?

Dropping the 2 in Apollonius

The theorem is AB² + AC² = 2AD² + 2BD². Writing AD² + BD² on the right halves the answer's square.

Using the whole side where the half is needed

BD in the theorem is half of BC, because D is the midpoint. Put BC itself in only through the form with 2AB² + 2AC² − BC².

Drill 2 more on apollonius theorem: the length of a median

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The centroid divides each median in the ratio 2 : 1
Centroid on a median
$AG = \frac{2}{3}AD \qquad GD = \frac{1}{3}AD$
A median halves the area, and the centroid makes equal parts
Areas made by medians
$\text{ar}(ABD) = \frac{1}{2}\,\text{ar}(ABC) \qquad \text{ar}(GBC) = \frac{1}{3}\,\text{ar}(ABC)$
Apollonius theorem: the length of a median
Apollonius theorem
$AB^2 + AC^2 = 2(AD^2 + BD^2) \qquad AD = \frac{1}{2}\sqrt{2AB^2 + 2AC^2 - BC^2}$

Watch out for (6)

Reading 2 : 1 from the wrong end
→ The centroid divides each median in the ratio 2 : 1
Using whole medians when two medians cross at 90°
→ The centroid divides each median in the ratio 2 : 1
Taking the centroid triangles as quarters
→ A median halves the area, and the centroid makes equal parts
Thinking a median quarters the area
→ A median halves the area, and the centroid makes equal parts
Dropping the 2 in Apollonius
→ Apollonius theorem: the length of a median
Using the whole side where the half is needed
→ Apollonius theorem: the length of a median

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