Medians and the Centroid

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Learning goals

Construct a median by locating the midpoint of the opposite side
Prove that three medians meet using midsegments, similarity, and a unique division point
Calculate median lengths while distinguishing the whole segment from its two parts
Compare the balance interpretation for a uniform sheet with other material distributions

Join a vertex to the opposite midpoint

A median of a triangle is the segment from a vertex to the midpoint of the opposite side. For example, if a side is 14 cm long, its midpoint is 7 cm from each endpoint. Joining that midpoint to the opposite vertex makes a median. There is one median from each of the three vertices.

The equal ticks identify the opposite midpoint. No right angle is required.

The word midpoint describes where the median ends. It does not say that the median meets the side at a right angle or divides the vertex angle equally. Those are separate conditions.

Worked example 1 Construct the median from one vertex

Problem. Given a nondegenerate triangle $ABC$ , construct its median from $A$ using a compass and unmarked straightedge. Explain why the constructed segment is a median.

Solution. Construct the perpendicular bisector of $BC$ , using the construction from the first lesson of this chapter. Set the compass to the same radius from $B$ and from $C$ , greater than half of $BC$ . The two pairs of arcs meet at distinct points on opposite sides of $BC$ .

Draw the line through those arc intersections. It meets $BC$ at its midpoint $D$ , because the perpendicular-bisector construction bisects the segment.

Draw segment $AD$ . Its endpoints are the vertex $A$ and the midpoint of the opposite side, so it is the required median. The perpendicular bisector was a tool for finding $D$ ; it is not generally the median $AD$ .

Check your understanding

In triangle $PQR$ , point $M$ lies on side $QR$ . Which fact makes segment $PM$ a median?

QM=MR

PM\perp QR

PQ=PR

\angle QPM=\angle MPR

Two medians give a fixed division

The crossing of two medians divides each into a longer vertex part and a shorter midpoint part. For a median of length 12 cm, the parts will be 8 cm from the vertex and 4 cm to the side midpoint. The longer part contains two copies of the shorter part, so the whole contains three. We will prove why the same split occurs in every triangle.

Let $D$ be the midpoint of $BC$ and $E$ the midpoint of $CA$ . Draw medians $AD$ and $BE$ , meeting at $G$ . Join $D$ to $E$ .

Compare triangle AGB with triangle DGE. The midsegment DE corresponds to AB.

The midsegment theorem says that the segment joining two side midpoints is parallel to the remaining side and half as long. Thus $DE$ is parallel to $AB$ . That parallel gives equal angles in the two small triangles, and their angles at $G$ are vertical angles. Similarity transfers the factor of two from $AB$ and $DE$ to the median parts.

Why two medians divide each other in a two-to-one ratio#

First follow the argument with lengths. Suppose $AB$ is 18 cm. The midsegment $DE$ is then 9 cm. Parallel angles and the vertical angles at $G$ give similar triangles $AGB$ and $DGE$ . The large-to-small factor is $18/9$ , or 2, so the vertex parts of both medians are twice their midpoint parts. If $AD$ is 15 cm, its three ratio parts are each 5 cm, giving a 10 cm vertex part. Nothing in that reasoning depends on 18 or 15.

For the general proof, assume $A,B,C$ are not collinear. Because $D$ is strictly between $B,C$ , the line $AD$ puts $B$ and $C$ on opposite sides. The interior point $E$ of $AC$ is on the same side of $AD$ as $C$ . Thus segment $BE$ crosses $AD$ . Both segments lie in the triangle, and they cannot meet at a boundary endpoint, so their crossing $G$ is strictly inside both medians.

The midsegment theorem gives

DE\parallel AB.

It also gives

AB=2DE.

The parallel lines give congruent alternate interior angles $GAB$ and $GDE$ . The angles $AGB$ and $DGE$ are vertical angles. Therefore AA gives

\triangle AGB\sim\triangle DGE.

These are genuine triangles. The interior crossing $G$ is not on $AB$ , and the line $DE$ meets $AD$ only at $D$ , not at $G$ . Their corresponding sides have the same factor, so

\frac{AG}{GD}=2.

Similarly,

\frac{BG}{GE}=2.

All four lengths are positive because $G$ is strictly inside the two medians. On $AD$ , segment addition now gives

\begin{aligned} AD&=AG+GD\\ AD&=3GD. \end{aligned}

The vertex part is twice the shorter part, not twice the whole median.

Check your understanding

In the proof configuration, why are triangles AGB and DGE similar?

A All medians have equal lengths B A median is perpendicular to its opposite side C Parallel angles and vertical angles give AA D Both triangles have three equal sides

Why the third median meets them

Three or more lines are concurrent when they pass through one common point. The midpoint conditions force concurrence here; it is not an extra fact you must assume from the drawing.

Why the third median passes through the same point#

In the numerical proof above, a median of length 15 cm has just one interior point 10 cm from its vertex. Two different crossings cannot both occupy that position.

In general, let $F$ be the midpoint of $AB$ . Draw the third median $CF$ , and call its crossing with $AD$ point $H$ . The same opposite-side argument used for the first pair puts $H$ strictly inside both segments.

Apply the preceding proof to medians $AD$ and $CF$ , using midsegment $DF$ parallel to $AC$ . It gives

AH=2HD.

Segment addition gives

HD=\frac13 AD.

But the first crossing also satisfies

GD=\frac13 AD.

Both $G$ and $H$ lie on the same segment from $D$ toward $A$ , at the same distance from $D$ . There is only one such point, so $H$ is $G$ . The third median therefore passes through $G$ , and the repeated two-median proof also gives

CG=2GF.

Each median joins a vertex to its own opposite midpoint.

The centroid of a nondegenerate triangle is this common point of its three medians. It is always inside the triangle, including when the triangle has a right or obtuse angle. On every median, the order is vertex, centroid, side midpoint, and the ratio in that order is $2:1$ .

Use Back and Next to construct each median. Drag A or change the base angles; BC stays fixed at 6 units. Compare AG with GD before comparing the other two vertex-to-centroid and centroid-to-midpoint pairs. The lengths change while the ordered ratios illustrate the proof above.

The three medians keep the same ordered two-to-one division

Construction: complete construction

Scroll sideways to see the whole diagram.

Angle B Angle C Construction Try a case

Angle B = 55°. Angle C = 65°. Angle A = 60°. The triangle is acute. D is the midpoint of BC. E is the midpoint of CA. F is the midpoint of AB. The three medians meet at G. AG ≈ 3.45 units. GD ≈ 1.73 units. AG : GD = 2 : 1. BG : GE = 2 : 1. CG : GF = 2 : 1. G is inside the triangle.

D, E and F are the opposite side midpoints. Read each ratio from a vertex toward its midpoint, with G between them.

Worked example 2 Use uniqueness rather than the appearance of a crossing

Problem. In triangle $ABC$ , $D,E,F$ are the midpoints of $BC,CA,AB$ . The median $AD$ has length 21 cm. Let $G$ be the crossing of $AD$ and $BE$ , and $H$ the crossing of $AD$ and $CF$ . Use the two-median result to explain why $G$ and $H$ are the same point.

Solution. Applied to the first pair, the result gives

DG=7\text{ cm}.

Applied to the second pair, it gives

DH=7\text{ cm}.

Both points are on segment $DA$ . The point 7 cm along that segment from $D$ is unique, so the names $G$ and $H$ identify the same point. The preceding three-median figure shows the resulting configuration.

Check your understanding

Why is the uniqueness step needed after proving the two-median ratio?

A Any three segments must be concurrent B It identifies the two crossings on the same median C It proves that every median has the same length D It turns a median into an angle bisector

Keep the parts and the whole separate

If the shorter part of median $AD$ is $t$ , its vertex part is $2t$ and its whole length is $3t$ . Thus

AG=\frac23 AD.

The other part is

GD=\frac13 AD.

Count two ratio parts from the vertex and one more to the midpoint.

Worked example 3 Recover a whole median from its vertex part

Problem. Triangle $ABC$ has centroid $G$ , and $D$ is the midpoint of $BC$ . If $AG$ is 14 cm, find $GD$ and $AD$ .

Solution. The given 14 cm is two ratio parts, so one part is

\begin{aligned} GD&=14\div2\\ GD&=7\text{ cm}. \end{aligned}

Add the two portions to recover the whole.

\begin{aligned} AD&=14+7\\ AD&=21\text{ cm}. \end{aligned}

The 14 cm part touches the vertex. The 7 cm part touches the side midpoint.

Worked example 4 Check whether two given expressions are compatible

Problem. On median $PM$ , point $G$ is the centroid. The lengths, in centimeters, are

PG=3x+2.

The other portion is

GM=x+4.

Find $x$ and the whole median $PM$ . Check that both portions are positive.

Solution. The vertex portion is twice the midpoint portion.

\begin{aligned} 3x+2&=2(x+4)\\ x&=6. \end{aligned}

Substitute into each given expression.

PG=20\text{ cm}.

Also,

GM=10\text{ cm}.

Both are positive, and 20 is twice 10. Adding them gives

PM=30\text{ cm}.

The checked lengths have the required order and ratio.

Check your understanding

On median AD, G is the centroid. If GD is 5 cm, what is the length of AD?

A 10 cm B 7.5 cm C 2.5 cm D 15 cm

A balance point depends on the material

A uniform triangular sheet has its balance point at the centroid. Here uniform means the same mass in equally sized pieces everywhere in the thin sheet, with no extra weight attached. This describes material spread across the whole triangular region.

Imagine the sheet made from many narrow strips parallel to $BC$ . Each strip has its middle on median $AD$ : a parallel cross-section has endpoints at the same fraction along the two sides, so their midpoint follows that median. A uniform strip balances at its middle. This explains physically why the sheet’s balance point lies on $AD$ . The same reasoning puts it on the other medians, so it is their intersection.

The midpoints of parallel cross-sections lie along AD. This supports the physical balance interpretation.

This is an explanation of the ideal physical model, not a new proof from area formulas. A real balancing experiment supplies evidence and may be affected by thickness, uneven material, or the support. The earlier similarity proof established the exact geometric ratio independently.

Worked example 5 Decide which balance claim the model supports

Problem. A uniform thin triangular sheet and a frame made only from uniform wire have the same scalene triangular outline. For which object does the sheet-balance interpretation guarantee balance at the intersection of the medians? Explain the difference.

The objects have the same outline but place their material in different regions.

Solution. The guarantee applies to the uniform sheet. Its material fills the whole triangular region with the same mass per equally sized piece.

The wire frame places material only along the boundary. The parallel-strip explanation for a filled sheet does not describe that object. Its balance point can differ, so the outline alone does not justify using the centroid as its center of mass. A sheet with an added weight also falls outside the uniform-sheet assumption.

Check your understanding

What additional information makes the centroid balance interpretation apply to a triangular cutout?

A Its boundary is made of equal-thickness wire B Its material is uniformly distributed through the thin sheet C Its longest side is horizontal D Its centroid is equally far from the vertices

Medians and the Centroid

Learning goals

Join a vertex to the opposite midpoint

Two medians give a fixed division

Why the third median meets them

Keep the parts and the whole separate

A balance point depends on the material

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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