CDS Mathematics · Triangles

Medians and Apollonius Theorem

Apollonius' theorem gives a median from the three sides; in a right triangle, Pythagoras on a point of one leg does the same job; and a perpendicular turns the difference of two squared sides into a difference of two squared pieces.

All Triangles notes CDS Mathematics strategy

Why this matters

Fourteen PYQs, and six of them are HARD, more than on any other page. Almost none needs the median formula itself. Most put a point on a leg of a right triangle and ask for a combination of squares, which two Pythagoras equations settle; the rest subtract two Pythagoras equations across an altitude.

Concept 1 of 3: Apollonius' theorem

A median splits the base in half. Write Pythagoras in the two halves, with the altitude from the vertex as a shared leg, and add: the unknown altitude cancels and leaves the median in terms of the sides.

Definition

If $AD$ is the median to $BC$ (so $BD = DC = \dfrac a2$ ):

$AB^2 + AC^2 = 2(AD^2 + BD^2)$ ;
equivalently $m_a^2 = \dfrac{2b^2 + 2c^2 - a^2}{4}$ ;
adding the three: $m_a^2 + m_b^2 + m_c^2 = \dfrac34(a^2 + b^2 + c^2)$ ;
in a right triangle the median to the hypotenuse is half the hypotenuse.

Apollonius' theorem

AB^2 + AC^2 = 2\left(AD^2 + BD^2\right)

Worked example

In triangle

ABC

AB = 5

AC = 7

and

BC = 8

. Find the median

AD

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (I) 2025 — Elementary Mathematics · Q74Hard

Example 1 · Triangles · Medians and Apollonius Theorem

The sides of a triangle are k, 1.5k and 2.25k. What is the sum of the squares of its medians ?

Half the base, not the base

Apollonius uses

BD^2 = \left(\dfrac{BC}{2}\right)^2

. Putting

BC^2

in its place is the usual slip, and it gives a median that is too short.

Drill 1 more on apollonius' theorem

Concept 2 of 3: Points on a leg of a right triangle

If the right angle is at

B

and

P

is any point on

BC

, then

ABP

is itself a right triangle. So every line from

A

to a point of

BC

gives one Pythagoras equation, and two such lines give two equations in the two legs.

Definition

With the right angle at $B$ , $AB = c$ , $BC = a$ :

for $P$ on $BC$ : $AP^2 = c^2 + BP^2$ ;
with $Q$ the midpoint of $BC$ and $P$ the midpoint of $AB$ : $AQ^2 = c^2 + \dfrac{a^2}{4}$ and $CP^2 = a^2 + \dfrac{c^2}{4}$ , so $AQ^2 + CP^2 = \dfrac54 AC^2$ ;
right angle at $C$ , $P$ on $AC$ and $Q$ on $BC$ : $AQ^2 + BP^2 = AB^2 + PQ^2$ .

Two midpoints of the legs

AQ^2 + CP^2 = \tfrac54\,AC^2

Worked example

ABC

is right-angled at

B

and

P

is the midpoint of

BC

. If

AP = 5

and

AC = \sqrt{73}

, find

AB

and

BC

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2026 · CDS (I) 2026 — Elementary Mathematics · Q55Moderate

Example 2 · Triangles · Medians and Apollonius Theorem

ABC is a triangle right-angled at C. Let P be the midpoint of BC. If

AP = 4\sqrt{13}

cm and

AB = 20

cm, then what is the perimeter of the triangle ABC ?

Measure from the right angle

AP^2 = AB^2 + BP^2

uses the piece of the leg from the right angle

B

P

. With

M

and

N

trisecting

BC

BN

is two-thirds of

BC

, not one-third.

Drill 6 more on points on a leg of a right triangle

Concept 3 of 3: Subtracting across an altitude

Drop the altitude

AD

BC

and write Pythagoras on both sides of it. Both equations contain

AD^2

, so subtracting them removes it, leaving two sides against the two pieces of the base.

Definition

Any triangle, $AD \perp BC$ : $AB^2 - AC^2 = BD^2 - DC^2$ . Factorised: $(AB + AC)(AB - AC) = BC\,(BD - DC)$ .
Isosceles, $AB = AC$ , any $D$ on $BC$ : $AB^2 - AD^2 = BD\cdot DC$ . (Drop the altitude to the midpoint $M$ and use $BM^2 - MD^2 = (BM - MD)(BM + MD)$ .)
Foot of an altitude: with $BD + DC = a$ and $BD - DC = \dfrac{c^2 - b^2}{a}$ , each piece follows.

Across an altitude

AB^2 - AC^2 = BD^2 - DC^2

Worked example

In triangle

ABC

AB = 13

AC = 15

and

BC = 14

AD

is perpendicular to

BC

. Find

BD

DC

and

AD

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q77Hard

Example 3 · Triangles · Medians and Apollonius Theorem

The perpendicular AD on the base BC of a triangle ABC intersects BC at D so that DB = 3 CD. Which one of the following is correct ?

The general form needs a perpendicular

AB^2 - AC^2 = BD^2 - DC^2

holds only when

AD

is the altitude. The isosceles form

AB^2 - AD^2 = BD\cdot DC

is different: there

D

can be any point of

BC

Drill 4 more on subtracting across an altitude

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)

Apollonius' theorem
Apollonius' theorem
$AB^2 + AC^2 = 2\left(AD^2 + BD^2\right)$
Points on a leg of a right triangle
Two midpoints of the legs
$AQ^2 + CP^2 = \tfrac54\,AC^2$
Subtracting across an altitude
Across an altitude
$AB^2 - AC^2 = BD^2 - DC^2$

Watch out for (3)

Half the base, not the base
→ Apollonius' theorem
Measure from the right angle
→ Points on a leg of a right triangle
The general form needs a perpendicular
→ Subtracting across an altitude

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