▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Drag A, B or C; the midpoints, G, P and Q follow, and the readout under the canvas keeps both ratios at 2:1. Learn more about why it holds
Proof?The medians AD and BE meet at G. Mark the midpoints P of AG and Q of BG. DE is a midsegment of triangle ABC and PQ a midsegment of triangle GAB, so both are parallel to AB and half as long: PQDE is a parallelogram, and its diagonals bisect each other at G. So AP = PG = GD and BQ = QG = GE: G is two thirds of the way down each median, and the third median passes through it too. Drag any vertex.Medians and the centroid
Display
Drag A, B or C; the midpoints, G, P and Q follow
AG : GD = 2.00 : 1, BG : GE = 2.00 : 1
Step 1 / 7
Step explanations
1Two medians meet at G
Given D and E are the midpoints of BC and CA, so AD and BE are medians of triangle ABC; they meet at G. To prove: AG = 2 GD and BG = 2 GE, and that the third median CF passes through G as well.
Step through the proof, then drag the vertices and watch the three medians keep meeting at one point, two thirds of the way down each of them. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
Proof?Medians and the centroidDisplayLabelsMeasuresDrag A, B or C; the midpoints, G, P and Q follow
AG : GD = 2.00 : 1, BG : GE = 2.00 : 1
← Back▶ PlayNext →ResetNormal ▾Step 1 / 7
Step explanations
1Two medians meet at G
Given D and E are the midpoints of BC and CA, so AD and BE are medians of triangle ABC; they meet at G. To prove: AG = 2 GD and BG = 2 GE, and that the third median CF passes through G as well.
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1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names
3Draggable points — A, B and C move freely; the hollow midpoints D, E, the crossing point G and later P, Q, F follow
4Readout — the line under the figure: the ratios AG : GD and BG : GE, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 7 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 7 of 7
Proof?Medians and the centroidDisplayLabelsMeasuresDrag A, B or C; the midpoints, G, P and Q follow
AG : GD = 2.00 : 1, BG : GE = 2.00 : 1
← Back↻ ReplayNext →ResetNormal ▾Step 7 / 7
Step explanations
1Two medians meet at G
Given D and E are the midpoints of BC and CA, so AD and BE are medians of triangle ABC; they meet at G. To prove: AG = 2 GD and BG = 2 GE, and that the third median CF passes through G as well.
2The midpoints P of AG and Q of BG
Construction Mark P, the midpoint of AG, and Q, the midpoint of BG. Both are hollow: they follow G.
3DE ∥ AB and DE = ½ AB
Midsegment theorem In triangle ABC, D and E are the midpoints of BC and CA, so DE is a midsegment: parallel to the third side AB and half as long.
4PQ ∥ AB and PQ = ½ AB
Midsegment theorem In triangle GAB, P and Q are the midpoints of GA and GB, so PQ is a midsegment of that triangle: also parallel to AB and half as long.
5PQDE is a parallelogram, so G is the midpoint of PD and of QE
Parallelogram diagonals PQ and DE are equal and parallel, so PQDE is a parallelogram. Its diagonals are PD, which lies along the median AD, and QE, which lies along the median BE; they cross at G, and the diagonals of a parallelogram bisect each other: PG = GD and QG = GE.
6AG = 2 GD and BG = 2 GE
Adding the parts AP = PG because P is the midpoint of AG, and PG = GD was just shown, so the median AD is cut into three equal parts and AG = AP + PG = 2 GD. In the same way BQ = QG = GE, so BG = 2 GE.
7The third median passes through G
Same argument Run the same argument with the medians AD and CF: they meet at the point two thirds of the way from A to D. That point is G, because G is the only point of AD with AG = 2 GD. So all three medians pass through G, the centroid, and each is divided there in the ratio 2 : 1. ∎
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1The centroid G — where all three medians meet, two thirds of the way down each
2The parallelogram PQDE — its diagonals PD and QE lie along two medians and bisect each other at G
3The third median CF — added at the last step; it passes through G as the argument predicts
4Justification tags — Given, Construction, Midsegment theorem (twice), Parallelogram diagonals, Adding the parts, Same argument
5Replay — at the last step Play becomes Replay and Next is off
The Statement
A median of a triangle joins a vertex to the midpoint of the opposite side. Every triangle has three, and the theorem says two things about them: they all pass through one point G, the centroid, and that point cuts each median in the same ratio,
AG:GD=BG:GE=CG:GF=2:1
two thirds of the median lie on the vertex side of G, one third on the side of the midpoint. Three lines meeting at one point is never an accident; the proof has to explain why the third median cannot miss the point where the first two cross.
The proof here follows the plan of the earlier pages: two midsegments that are both parallel to AB and half of it make a parallelogram, and the diagonals of that parallelogram are pieces of the medians. In the player the midpoints D, E, F, the crossing point G and the construction points P, Q are hollow, computed from the vertices. The readout under the canvas shows the two ratios AG:GD and BG:GE, live while you drag.
Opening scene, frozen
The medians AD and BE meeting at the hollow point G.
Two New Midpoints
Start with two of the medians, AD and BE, and their crossing point G. The construction marks the midpoint P of AG and the midpoint Q of BG, each with the same tick as the rest of its median: one tick on the pieces of AD, two on the pieces of BE.
The idea is to look at the small triangle GAB, which has the same base AB as the big triangle and its apex at G. The midpoints P and Q belong to this small triangle in the way D and E belong to the big one, so each triangle gives a midsegment parallel to AB. Comparing the two midsegments is what pins G down.
Two new midpoints, frozen
P halfway along AG and Q halfway along BG, with the ticks of their medians.
The First Midsegment
In triangle ABC, D and E are the midpoints of BC and CA, so DE is a midsegment, and the midsegment theorem gives
DE∥AB,DE=21AB
The player fills the big triangle and marks DE and AB with one chevron each. Notice where D and E sit: at the far ends of the two medians, below G.
The first midsegment, frozen
DE in the big triangle: parallel to AB, half of it.
The Second Midsegment
In the small triangle GAB, P and Q are the midpoints of GA and GB, so PQ is a midsegment of that triangle, and the same theorem gives
PQ∥AB,PQ=21AB
Now DE and PQ are both parallel to AB and both equal to half of it. So they are parallel to each other and equal in length, and they sit on opposite sides of G: P and Q on the vertex side of the medians, D and E on the far side.
The second midsegment, frozen
PQ in the small triangle GAB: also parallel to AB and half of it.
A Parallelogram at G
A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram; the midsegment page proved this on the way to its own theorem, with a diagonal and a congruence, and the opposite sides page gives the converse direction. So PQDE is a parallelogram, shown filled in the player.
Its diagonals are PD and QE. But P, G, D all lie on the median AD, so the diagonal PD is a piece of that median, and likewise QE is a piece of the median BE; the diagonals cross at G. The diagonals of a parallelogram bisect each other, so
PG=GD,QG=GE
The parallelogram at G, frozen
PQDE filled; its diagonals PD and QE lie along the medians and bisect each other at G.
Two Thirds of the Way
Put the pieces of the median AD side by side. AP=PG because P is the midpoint of AG, and PG=GD from the parallelogram. So the three pieces AP, PG, GD are equal, the median is cut into thirds, and
AG=AP+PG=2⋅GD
The same on the median BE: BQ=QG=GE, so BG=2⋅GE. The crossing point of any two medians is two thirds of the way down each of them, from the vertex. The player lights the six equal pieces.
Two thirds of the way, frozen
AP = PG = GD and BQ = QG = GE: each median cut into thirds.
Why It Holds
Nothing in the argument depended on which two medians were chosen. Take the medians AD and CF instead: they meet at a point which, by the same proof, is two thirds of the way from A to D. But there is only one such point on AD, and it is G. So CF passes through G as well: the three medians are concurrent, and G divides each of them in the ratio 2:1. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given medians, a construction, the midsegment theorem twice, the parallelogram and its diagonals, an addition, and the same argument once more.
The point G is called the centroid. It is the triangle's balance point: a cardboard triangle rests level on a pin placed there, because each median splits the triangle into two halves of equal area, and the three medians together split it into six small triangles of equal area.
The conclusion, frozen
The third median CF drawn: it passes through G too.
Dragging the Figure
Drag any vertex: the midpoints slide, the medians swing, and the readout keeps both ratios at exactly 2:1. Pull A far to one side and the triangle leans over; the centroid leans with it but never leaves the inside of the triangle, because it is always two thirds of the way along a median that lies inside.
Watch the parallelogram PQDE as you drag: it changes shape but stays a parallelogram, with G at the crossing of its diagonals. A drag that would flatten the triangle, or carry a vertex out of the frame, is refused.
After dragging A, frozen
A pulled right: the triangle leans, the medians swing, and G is still two thirds of the way down each.
Consequences
• Coordinates. The centroid is the average of the three vertices, G=31(A+B+C), which is the 2:1 ratio written with vectors: G=D+31(A−D) and D=21(B+C). • Centre of mass. Three equal weights at the vertices, or a uniform triangular plate, balance at the centroid; the plate's balance point is the same as the three weights' because each median divides the area in half. • Six equal triangles. The three medians cut the triangle into six triangles of equal area, each one sixth of the whole. • The medial triangle. The triangle DEF of the midpoints has the same centroid as ABC, and so does the midpoint parallelogram of Varignon's theorem for any quadrilateral, at the crossing of its diagonals. • The Euler line. The centroid lies on the line through the circumcentre and the orthocentre, twice as far from the orthocentre as from the circumcentre.