Medians of a Triangle

A median of a triangle is the segment that joins a vertex with the midpoint of the opposite side.

In the figure $A_1$ is the midpoint of $BC$ (${BA_1=A_1C}$), and $AA_1$ is the median from the vertex $A.$ Its length is written $m_a,$ the median to the side $a.$ Every triangle has three medians, $AA_1,$ $BB_1$ and $CC_1,$ with lengths $m_a,$ $m_b$ and $m_c.$ Unlike the altitudes, the medians always lie inside the triangle.

On this page: Centroid · Medians and area · Length of a median · Special triangles · Solved problems · Common mistakes · Practice

The Centroid

In other words, the centroid lies two thirds of the way from each vertex to the midpoint of the opposite side:

${AG=\frac23\,m_a},\qquad{GA_1=\frac13\,m_a}.$

If you cut a triangle out of cardboard, it balances on the tip of a pencil placed at the centroid. That is why the centroid is also called the center of gravity of the triangle.

Why do the medians meet at one point, and why is the ratio ${2:1}?$

Let the medians $BB_1$ and $CC_1$ meet at $G.$ The segment $B_1C_1$ joins the midpoints of $AC$ and $AB,$ so it is a midsegment: ${B_1C_1\parallel BC}$ and ${B_1C_1=\frac12\,BC}.$

Triangles $GB_1C_1$ and $GBC$ have equal angles: alternate angles at the parallel lines $B_1C_1$ and $BC,$ and vertical angles at $G.$ So they are similar with ratio ${\frac{B_1C_1}{BC}=\frac12}.$ Hence ${GB_1=\frac12\,GB}$ and ${GC_1=\frac12\,GC}:$ the point $G$ divides each of the medians $BB_1$ and $CC_1$ in the ratio ${2:1}.$

The same argument for the medians $AA_1$ and $BB_1$ shows that they meet at the point of $BB_1$ that divides it in the ratio ${2:1}$ from $B.$ That is the same point $G,$ so all three medians pass through $G.$

The centroid in coordinates. If the vertices are ${A(x_1,\ y_1)},$ ${B(x_2,\ y_2)}$ and ${C(x_3,\ y_3)},$ the centroid is the point

${G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)}.$

Medians and Area

A median divides a triangle into two triangles with equal areas.

The triangles $ABA_1$ and $AA_1C$ have equal bases, ${BA_1=A_1C},$ and the same altitude from $A,$ so their areas are equal.

The three medians divide the triangle into six small triangles with equal areas. Each of them is ${\frac16}$ of the area of $ABC,$ and, for example, ${S_{ABG}=\frac13\,S_{ABC}}.$

Why are the six areas equal?

The triangles $GBA_1$ and $GA_1C$ have equal bases and a common altitude from $G,$ so their areas are equal; call this area $x.$ The median $AA_1$ divides $ABC$ into two halves, so ${S_{ABG}}={\frac12\,S_{ABC}-x}={S_{ACG}}.$ In the same way ${S_{ABG}=S_{BCG}}.$ So the triangles $ABG,$ $BCG$ and $CAG$ have equal areas, each ${\frac13\,S_{ABC}},$ and each of them is divided by a median into two equal halves of ${\frac16\,S_{ABC}}.$

The Length of a Median

In each formula the side to which the median is drawn comes with a minus sign.

A proof with a parallelogram

Extend the median $AA_1$ beyond $A_1$ by its own length to a point $D,$ so that ${A_1D=AA_1=m_a}.$ The diagonals $AD$ and $BC$ of the quadrilateral $ABDC$ bisect each other at $A_1,$ so $ABDC$ is a parallelogram with sides $b$ and $c$ and diagonals ${AD=2m_a}$ and ${BC=a}.$

In every parallelogram the sum of the squares of the diagonals equals the sum of the squares of the four sides:

${(2m_a)^2+a^2=2b^2+2c^2}.$

So ${4m_a^2=2b^2+2c^2-a^2},$ which is the formula. The parallelogram rule itself follows from the law of cosines for the two diagonals.

Sides from the medians. If the three medians are known, the sides are

Adding the squares of the three median formulas gives one more neat relation:

${m_a^2+m_b^2+m_c^2}={\frac34\left(a^2+b^2+c^2\right)}.$

Medians of Special Triangles

• Isosceles triangle. The median to the base is also the altitude to the base and the bisector of the vertex angle. The medians to the two legs are equal.
• Equilateral triangle. The medians are equal to the altitudes, ${m=\frac{a\sqrt3}{2}},$ and the centroid is the center of the triangle.
• Right triangle. The median to the hypotenuse is half the hypotenuse: ${m_c=\frac{c}{2}}.$ The midpoint of the hypotenuse is equally far from the three vertices, so it is the center of the circumscribed circle.

Conversely, if a median is half of the side to which it is drawn, then the angle opposite this side is a right angle.

Why?

Let ${CC_1=\frac12\,AB}.$ Then ${C_1A=C_1C=C_1B},$ so the triangles $AC_1C$ and $BC_1C$ are isosceles, and ${\angle C_1CA=\alpha},$ ${\angle C_1CB=\beta}.$ So ${\gamma=\alpha+\beta},$ and from ${\alpha+\beta+\gamma=180^\circ}$ we get ${\gamma=90^\circ}.$

Solved Problems

Problem 1. In triangle $ABC$ the sides are ${a=10}$ and ${b=c=13}.$ Find the medians $m_a$ and $m_b.$

Solution.

${m_a}={\frac12\sqrt{2\cdot 169+2\cdot 169-100}}={\frac12\sqrt{576}}={12},$

${m_b}={\frac12\sqrt{2\cdot 100+2\cdot 169-169}}={\frac12\sqrt{369}}={\frac{3\sqrt{41}}{2}}\approx{9.60}.$

Check $m_a$ another way: the triangle is isosceles with base $a,$ so $m_a$ is also the altitude to the base, ${\sqrt{13^2-5^2}=12}.$

Problem 2. The legs of a right triangle are ${a=6}$ and ${b=8}.$ Find the median to the hypotenuse and the median to the leg $a.$

Solution. The hypotenuse is ${c=10},$ so ${m_c=5}.$ The median to the leg $a$ goes from $A$ to the midpoint of $BC.$ It is the hypotenuse of a right triangle with legs $b$ and ${\frac a2},$ so

${m_a}={\sqrt{b^2+\left(\frac a2\right)^2}}={\sqrt{64+9}}={\sqrt{73}}\approx{8.54}.$

The formula gives the same: ${\frac12\sqrt{2\cdot 64+2\cdot 100-36}}={\frac12\sqrt{292}}={\sqrt{73}}.$

Problem 3. The median $AA_1$ is 15 cm. Find $AG$ and $GA_1.$

Solution. ${AG=\frac23\cdot 15=10}$ cm and ${GA_1=\frac13\cdot 15=5}$ cm.

Problem 4. Find the centroid of the triangle with vertices ${A(1,\ 2)},$ ${B(5,\ 0)}$ and ${C(3,\ 7)}.$

Solution. ${G\left(\frac{1+5+3}{3},\ \frac{2+0+7}{3}\right)}={G(3,\ 3)}.$

Problem 5. The area of triangle $ABC$ is 36 cm². Find the areas of the triangles $ABA_1,$ $ABG$ and $GBA_1.$

Solution. ${S_{ABA_1}=\frac12\cdot 36=18},$ ${S_{ABG}=\frac13\cdot 36=12}$ and ${S_{GBA_1}=\frac16\cdot 36=6}$ (all in cm²).

Problem 6. The medians of a triangle are ${m_a=9},$ ${m_b=12}$ and ${m_c=15}.$ Find its sides.

Solution.

${a}={\frac23\sqrt{2\cdot 144+2\cdot 225-81}}={\frac23\sqrt{657}}={2\sqrt{73}}\approx{17.09},$

${b}={\frac23\sqrt{2\cdot 81+2\cdot 225-144}}={\frac23\sqrt{468}}={4\sqrt{13}}\approx{14.42},$

${c}={\frac23\sqrt{2\cdot 81+2\cdot 144-225}}={\frac23\sqrt{225}}={10}.$

Problem 7. Prove that a triangle with two equal medians is isosceles.

Solution. If ${m_a=m_b},$ then ${2b^2+2c^2-a^2}={2a^2+2c^2-b^2},$ so ${3b^2=3a^2}$ and ${a=b}.$

Common Mistakes

Mistake	Wrong	Right
The ratio taken from the wrong end	${AG:GA_1=1:2}$	${AG:GA_1=2:1};$ the longer part is next to the vertex
Median and altitude mixed up	The median is perpendicular to the opposite side	That is the altitude. A median is perpendicular to its side only when it is drawn to the base of an isosceles triangle
Median and angle bisector mixed up	The median from $A$ halves the angle at $A$	That is the angle bisector. In a scalene triangle the median, the bisector and the altitude from a vertex are three different segments
A wrong sign in the formula	${m_a=\frac12\sqrt{2b^2+2c^2+a^2}}$	${m_a=\frac12\sqrt{2b^2+2c^2-a^2}}$
The median to a leg of a right triangle	Legs 6 and 8, so the median to the leg 6 is half of it, 3	${\sqrt{8^2+3^2}=\sqrt{73}};$ only the median to the hypotenuse is half of its side

Practice Problems

Try them yourself, then open the answer.

1. The median $AA_1$ is 12. Find $AG$ and $GA_1.$

${AG=8}$ and ${GA_1=4}.$

2. ${GA_1=3}.$ Find the median $AA_1.$

${AA_1=3\cdot 3=9}$

3. The hypotenuse of a right triangle is 17. Find the median to the hypotenuse.

${\frac{17}{2}=8.5}$

4. The sides of a triangle are ${a=8}$ and ${b=c=5}.$ Find $m_a.$

${m_a}={\frac12\sqrt{50+50-64}}={\frac12\cdot 6}={3}.$ It is also the altitude to the base: ${\sqrt{25-16}=3}.$

5. The sides of a triangle are 5, 6 and 7. Find the median to the side 6.

${\frac12\sqrt{2\cdot 25+2\cdot 49-36}}={\frac12\sqrt{112}}={2\sqrt7}\approx{5.29}$

6. Find the centroid of the triangle with vertices ${(0,\ 0)},$ ${(6,\ 0)}$ and ${(3,\ 9)}.$

${G(3,\ 3)}$

7. The area of a triangle is 30. Find the area of each of the six small triangles formed by its medians.

${30:6=5}$

8. Find the median of an equilateral triangle with side 6 and the distance from the centroid to a vertex.

${m=\frac{6\sqrt3}{2}=3\sqrt3\approx 5.20}$ and ${AG}={\frac23\cdot 3\sqrt3}={2\sqrt3}\approx{3.46}.$

9. The sides of a triangle are 4, 5 and 6. Find ${m_a^2+m_b^2+m_c^2}.$

${\frac34\,(16+25+36)}={\frac34\cdot 77}={57.75}$

10. In triangle $ABC$ the median ${m_c=5},$ the side ${c=10}$ and the side ${a=6}.$ Find the side $b.$

${m_c=\frac c2},$ so ${\gamma=90^\circ}$ and ${b=\sqrt{10^2-6^2}=8}.$

More about triangles: Triangles · Altitude · Midsegment · Congruence