## How three medians meeting at centroid divides triangle area into 3 and 6 equal parts and properties of cevians

Centroid divides triangle area into 3 equal parts formed by the longer median segments at centroid and 6 equal parts by all six median segments at centroid.

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Also a point on a cevian divides a triangle area in the ratio of its segments at the point.

Medians meeting at centroid create a rich set of *relations involving fragmented areas and segmented lines including the sides of the triangle.*

Contents are,

**How each median divides triangle area into two equal parts.**Relation between areas of the triangles formed by the medians at centroid and a line parallel to a base and passing through the mid-points of the other two sides.**Area to cevian segment ratio concept:** A triangle area is divided in the same ratio as the cevian segment ratio at a point that acts as the vertex of a second triangle.**Area of triangle** from **length of medians.**

### A median divides a triangle into 2 parts of equal area

In the following figure, AD is a median of $ riangle ABC$ bisecting the opposite side BC at D. The centroid is the point G through which the other two medians, if drawn, would pass. AH is the altitude of the triangle with BC as the base; and PAQ is a line parallel to base BC.

With base BC and altitude AH the area of $ riangle ABC$ is,

$A=frac{1}{2}BC imes{AH}$.

AS $BD=CD=frac{1}{2}BC$, area of $ riangle ABD$ is,

$A_{ABD}=frac{1}{2}BD imes{AH}=frac{1}{4}BC imes{AH}=frac{1}{2}A$.

For $ riangle ADC$ also altitude is same AH, like any triangle with base BC and vertex lying on line PAQ parallel to it. So area of $ riangle ADC$ is,

$A_{ADC}=frac{1}{2}DC imes{AH}=frac{1}{4}BC imes{AH}=frac{1}{2}A$.

Thus the median AD divides the area of the triangle into two equal parts.

Before we next show you how three medians divide the area of a triangle into six equal parts, we will generalize the first result into the powerful concept of * Area to base division ratio*, and show its mechanism.

### Area to base division ratio concept

The following figure represents the problem solution.

In $ riangle ABC$, as line segment AD from vertex A to base BC divides BC in the ratio of, $CD:BD=x:y$, the ratio of areas of $ riangle ACD$ and $ riangle ABD$ will also be $x:y$. To formally state this important general result,

A line segment from a vertex of a triangle to the opposite side, which is the base, divides the base and the triangle area in the same ratio.

Proof of area to base division ratio concept

In $ riangle ABC$, line segment AD divides the base BC at D so that,

$CD:BD=x:y$.

As both the triangles, $ riangle ACD$ and $ riangle ABD$ have same altitude AP which is also the altitude of parent $ riangle ABC$,

$ ext{Area of } riangle ABD=A_{ABD}=frac{1}{2}BD imes{AP}$.

Similarly,

$ ext{Area of } riangle ACD=A_{ACD}=frac{1}{2}CD imes{AP}$.

Taking the ratio of the latter to the former,

$A_{ACD}:A_{ABD}=CD:BD=x:y$.

The result is really simple but general.

A special case is when AD is the median and $CD=BD$. In this case then, the median divides the triangle into two parts of equal areas.

As another example, if $BD=2CD$,

$A_{ABD}=2A_{ACD}$.

We will use this last result to explain how the three medians divide a triangle into six regions of equal area.

### All three medians together divide a triangle into 6 equal parts, proof

The following figure will help explain the mechanism of this relationship. AD, BE and CF are the three medians dividing the $ riangle ABC$ into six part triangles meeting point of which is the centroid G. AP is the perpendicular to median BE and is the altitude of all three triangles $ riangle ABG$, $ riangle AEG$ and $ riangle ABE$.

The medians divide the triangle into six non-overlapping triangular regions with vertices meeting at centroid G. These six triangles actually consist of three pairs of equal triangles formed by the division of three triangles formed from the centroid and three pairs of vertices, by the part of median from vertex G to the base. For example, areas of one such equal area pair of triangles hold the relation,

$A_{BGD}=A_{CGD}$.

Our objective is to show area of one adjacent pair of the three pairs of triangles as equal. For example we will show,

$A_{AFG}=A_{AEG}$.

$ riangle ABE$ is divided into two triangles $ riangle ABG$ and $ riangle AEG$ by the median section $AG$ incident on base BE at G.

As BE is a median, by the * median section ratio at centroid* concept, it is divided into two sections BG and EG with a ratio, $BG:EG=2:1$ .

So by the * area to base division ratio* concept,

$A_{ABG}=2A_{AEG}$.

Again, in $ riangle ABG$,

$A_{ABG}=2A_{AFG}$.

Thus,

$A_{AEG}=A_{AFG}$, that is, a pair of adjacent triangles are equal in area.

This makes areas of all six triangles with coincident vertices at G equal.

### Three vertex to centroid line segments divide a triangle into 3 equal parts

By this concept, G being the centroid, in figure below GA, GB and GC divide the triangle into 3 equal parts.

This can easily be concluded by adding up adjacent equal pairs of six equal triangles created by the three medians as in the previous section.

Before we proceed to the next sections we need to explain the often used **Triangle similarity rich concept.**

### Triangle similarity rich concept, the mechanism

The following figure will aid explanation of the concept.

We formally state the * triangle similarity rich concept* as,

Ratio of all pairs of corresponding sides of corresponding pairs of triangles formed by a straight line parallel to the base will be equal.

* Alternatively *this result leads to,

A straight line parallel to the base of a triangle will divide all straight line sections dropped from the vertex to the base in equal ratio.

This powerful rich concept is used frequently in both forms.

To be specific, with respect to the figure above, the line $PQ||BC$, the base, cuts across four lines $AB$, $AM$, $AN$ and $AC$ dropped from vertext A to the base BC at the points D, E, F and G respectively in equal ratio. That is to say,

$AD:BD=AE:EM=AF:FN=AG:GC$.

* This is the second form of the rich concept* and it follows from the broader first form of the definition of the concept.

Primary form of the triangle similarity rich concept

The line PQ||BC forms six pairs of corresponding triangles by cutting across the four lines dropped from the vertex to the base,

$ riangle ADE$ and $ riangle ABM$,

$ riangle AEF$ and $ riangle AMN$,

$ riangle AFG$ and $ riangle ANC$,

$ riangle ADF$ and $ riangle ABN$,

$ riangle AEG$ and $ riangle AMC$, and finally,

$ riangle ADG$ and $ riangle ABC$.

In each such pair of triangles, the ratio of the corresponding sides will be equal, so that ratio of all pairs of corresponding sides will be equal.

This amounts partially to,

$displaystylefrac{AD}{AB}=frac{AE}{AM}=frac{AF}{AN}=frac{AG}{AC}=frac{DG}{BC}$.

There will be more such equal ratios.

This happens primarily because, say in the pair of triangles, $ riangle ADG$ and $ riangle ABC$,

the angle at the vertex $angle A$ is common, andthe rest of the two pairs of angles, $angle ADG=angle ABC$, and $angle AGD =angle ACB$, as each of AB and AC intersects a pair of parallel lines DG and BC.

The three pairs of corresponding angles being equal, the $ riangle ADG$ and $ riangle ABC$ are similar so that ratio of all three pairs of corresponding sides become equal.

The same mechanism works in every pair of corresponding triangles.

It is trivial to show that this result leads to the second form of definition of this important rich concept.

We need to explain one more rich concept, the * area to cevian segment ratio*, before taking up the next sections.

A **cevian** is,

A line from a vertex reaching or crossing the opposite side of a triangle.

### Area to cevian segment ratio concept, proof

The following figure will aid the definition and proof of concept.

A * cevian is defined* as,

A line from a vertex reaching or crossing the opposite side of a triangle.

In figure above AD is such a cevian. * Special cases of cevians* are, the

*and the*

**median***of the triangle.*

**altitude*** Area to cevian segment ratio concept* formally states,

Any point on a cevian dividing the line into a ratio of $x:y$ will also divide the whole area of the triangle into two regions in the same ratio by acting as the vertex of second triangle inside the main triangle.

Alternately,

Ratio of the two line sections made by a point on a cevian will be same as the ratio of the two areas formed by the point as a vertex of a triangle with base same as the original triangle.

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Specifically with respect to the figure,

$AF:FD= ext{Area of region }ABFCA: ext{Area of } riangle BFC$.

Here point F divides the cevian AD into two sections FD and AF at a ratio of $x:y$.

It follows from this result,

$FD:AD=A_{AFD}:A_{ABC}=x:(x+y)$.

We get this result just by fraction inversion, addition of 1 and then inversion back.

Let us see why this concept works.

In $ riangle ABD$ with AD as base, applying * area to base division ratio concept*,

$A_{DBF}:A_{ABF}=FD:FA=x:y$.

Similarly in $ riangle ACD$ with AD as base, and applying the same concept we get,

$A_{DCF}:A_{ACF}=FD:FA=x:y$.

So by ratio concepts,

$A_{BFC}:A_{ABFC}=x:y$,

Or, $A_{BFC}:A_{ABC}=x:(x+y)$.

Special case of centroid as the cevian point on a median, the cevian in this case

As a special case, if F is the centroid, AD is a median so that F divides the median in a ratio,

$FD:AF=1:2$, and so in this special case,

$A_{BFC}:A_{ABC}=1:3$.

In other words, the area of the $ riangle BFC$ will be one-third of the main $ riangle ABC$.

With these rich concepts in place we are now ready to go ahead in dealing with the rest of the concepts elegantly.

### Additional useful relations between the areas of the triangles formed by the medians at the centroid

We will use the following figure to explain the concept in this section. Specifically we will see how the area of the triangles $ riangle GFP$, $ riangle GEP$ or $ riangle BGC$ are related to each other and to the area of the main $ riangle ABC$.

In this section we will explore what are the ratios of the areas of the triangles, $ riangle GFP$, $ riangle GFE$ and $ riangle BGC$ with respect to the area of the parent triangle.

Area of $ riangle GFP$ with respec to area of $ riangle GFE$

As median AD is the line from vertex to the last triangle base BC, passing through the bases of triangles, $ riangle AFE$ and $ riangle AGE$, where these two have a common base FE, the median AD bisects FE also at P. We conclude this result by applying the Triangle similarity rich concept.

So,

$FP=EP$.

Thus GP divides the $ riangle GFE$ in two equal parts. We conclude this from the concept of a median bisecting the area of a triangle.

This result says then,

$A_{GFP}=frac{1}{2}A_{GFE}$.

Area of $ riangle GFE$ with respect to area of $ riangle ABC$

To solve this problem we will determine the extent of GP.

By * triangle similarity rich concept*,

$AP=DP=DG+PG$.

Adding PG to the equation,

$AP+PG=AG=DG+2PG$.

Again by * median section ratio at centroid concept*,

$AG=2DG=DG+2PG$.

So,

$DG=2PG$, which we suspected to be true, but now we know.

As $DG=frac{1}{3}AD$,

$PG=frac{1}{6}AD$, where $AD$ is the median.

Flipping $ riangle GFP$ vertically with its vertex moved along the median

When we do this, we form the rectangle GEHF with its diagonals bisected at P, and its area bisected by common base FE.

Also,

$PH=PG=frac{1}{6}AD=frac{1}{3}AP$.

By * area to cevian segment ratio concept* then,

$A_{GFE}=A_{FHE}=frac{1}{3}A_{AFE}=2A_{GFP}$,

Also,

$A_{AFE}=frac{1}{4}A_{ABC}$, as both base and altitude of the smaller triangle are half of those of the larger triangle.

Relating the areas of all these triangles,

$A_{GFP}=frac{1}{2}A_{GFE}=frac{1}{6}A_{AFE}=frac{1}{24}A_{ABC}=frac{1}{8}A_{BGC}$.

The last result we get fro an earlier result of,

$A_{BGC}=frac{1}{3}A_{ABC}$, as GD is one-third of AD.

**Note** that we have not used the altitude or the base lengths. We could avoid this because of symmetric relationship between the median and the bases and hence the areas through the use of the powerful rich concepts.

### Area of triangle from lengths of its medians, proof

The figure below will aid the explanation.

As usual, AD, BE and CF are the three medians of $ riangle ABC$ intersecting at centroid G.

The values, $AD=m_1$, $BE=m_2$ and $CF=m_3$ are given. We are to derive the area of the $ riangle ABC$.

As we know, given three sides of a triangle as a, b and c, its semi-perimeter,

$s=frac{1}{2}(a+b+c)$, and the area,

$A=sqrt{s(s-a)(s-b)(s-c)}$.

We need then to construct a triangle with three sides as the medians, derive relation between the area of triangle of medians to the area of the original triangle and then get the area of the original triangle by deriving area of triangle of medians by using semi-perimeter concept.

Construction of triangle of medians

To construct the triangle of medians, first we have kept the side of median AD fixed, and **translated** the side of median BE in direction parallel to BC through the distance BD to reach the position DQ. This parallel translation has resulted in, $BE||DQ$ and $BE=DQ$ as well as, $EQ||BD||BC$ and $EQ=BD=DC$. This has formed the second side of the triangle of medians.

To form the third side, the third median CF has been translated in direction parallel to AB to AQ, so that, $CF||AQ$ and $CF=AQ$ as well as, $CQ||AF||AB||DE$ and $CQ=AF=FB=DE$.

In resulting parallelogram DEQC, the two diagonals bisect each other, so that, $ER=CR$ and $DR=QR$, so that **AR is the median** of $ riangle ADQ$ that is made up of medians ot the original $ riangle ABC$.

Thus AR divides the area of $ riangle ADQ$ into two equal parts each of which, say, is $x$.

AS $AE=EC=2ER$, $AR=AE+ER=3ER$, that is,

$ER=frac{1}{3}AR$, giving

$A_{DER}=A_{DCR}=frac{1}{3}A_{ADR}=z$, say.

Now we will use the following figure simplified from the above figure to put forth the final piece of reasoning.

We have removed the medians BE and CF as a cleanup measure.

Let us denote,

Area of $ riangle ABD=A_{ABD}=y$,

Area of $ riangle ADR=A_{ADR}=x$, and

Area of $ riangle DCR=A_{DCR}=z$.

And area of $ riangle ABC$ is, $A_{ABC}=2y$, and area of $ riangle ADQ$ is, $A_{ADQ}=A_m=2x$.

Now,

$A_{ACD}=y=x+z=x+frac{1}{3}x=frac{4}{3}x$.

So,

$A_{ABC}=frac{4}{3}A_m$.

In other words, **the area of any triangle is four-thirds of the area of the triangle formed from its medians.**

As we know how to find the area of a triangle from its given side lengths, it is a simple step more to find the area of the $ riangle ABC$ from the given length of its medians.

Lastly we will just skim through the question of finding the area of an equilateral triangle in terms of its sides or medians.

### Area of an equilateral triangle from its sides and medians

The following figure will aid the explanation.

As median AD of equilateral $ riangle ABC$ of side length $a$ is perpendicular to opposite side BC bisecting it, the area of the triangle is,

$A_{ABC}=frac{1}{2}BC imes{AD}$.

By the use of Pythagorean theorem we have,

$AD=m=sqrt{a^2-left(frac{a}{2}

ight)^2}=sqrt{frac{3}{4}a^2}=frac{sqrt{3}}{2}a$.

So,

$A_{ABC}=frac{sqrt{3}}{4}a^2=frac{1}{sqrt{3}}m^2$, where $a$ is the side length of the equilateral $ riangle ABC$ and $m$ is the length of all its three medians.

**Note:** Being USERS of knowledge for solving problems the best way possible, we need to know the mechanism behind a concept in as clear terms as possible. The clarity of understanding of a concept goes a long way in increasing our belief on the concept and consequently our ability to use the concept when it is really needed.

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