## Area of a triangle has intimate relationship with its medians

We have already introduced the richness of concepts related to medians in our previous discourse, **Basic and rich Geometry concepts 5, proof of median relations.**

In this session we will go through another facet of the medians, namely how medians are related to the area of the parent triangle.

The whole session will be mainly divided into two parts,

- How medians divide the area of a triangle, and
- How area of a triangle can be derived from the known lengths of its medians.

Among various ways medians divide the area of a triangle we will discuss the mechanisms of how,

- Each median divides a triangle into two equal parts,
- The three medians together divide a triangle into six equal parts, and
- The three vertex to centroid line segments divide the area of a triangle into three equal parts.

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

In the following figure, AD is a median of $\triangle ABC$ 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 $\triangle ABC$ is,

$A=\frac{1}{2}BC\times{AH}$.

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

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

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

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

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 $\triangle ABC$, 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 $\triangle ACD$ and $\triangle ABD$ 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 $\triangle ABC$, line segment AD divides the base BC at D so that,

$CD:BD=x:y$.

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

$\text{Area of }\triangle ABD=A_{ABD}=\frac{1}{2}BD\times{AP}$.

Similarly,

$\text{Area of }\triangle ACD=A_{ACD}=\frac{1}{2}CD\times{AP}$.

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 six equal parts, proof

The following figure will help explain the mechanism of this relationship. AD, BE and CF are the three medians dividing the $\triangle ABC$ 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 $\triangle ABG$, $\triangle AEG$ and $\triangle ABE$.

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,

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

$\triangle ABE$ is divided into two triangles $\triangle ABG$ and $\triangle AEG$ 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 $\triangle ABG$,

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

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 the triangle into three equal parts

By this concept, G being the centroid, in figure below GA, GB and GC divide the triangle into three 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,

$\triangle ADE$ and $\triangle ABM$,

$\triangle AEF$ and $\triangle AMN$,

$\triangle AFG$ and $\triangle ANC$,

$\triangle ADF$ and $\triangle ABN$,

$\triangle AEG$ and $\triangle AMC$, and finally,

$\triangle ADG$ and $\triangle ABC$.

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,

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

There will be more such equal ratios.

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

- the angle at the vertex $\angle A$ is common, and
- the 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 $\triangle ADG$ and $\triangle ABC$ 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 bridge line section ratio*, before taking up the next sections.

### Area to bridge line section ratio concept, proof

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

We define a * bridge line* as,

Any line from vertex to the base of a triangle bridging the gap between the two is defined to be a

.bridge line

In figure above AD is such a bridge line. * Special cases of bridges lines* are, the

*and the*

**median***of the triangle.*

**altitude*** Area to bridge line section ratio concept* formally states,

Any point on a bridge line 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 bridge line 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.

Specifically with respect to the figure,

$AF:FD=\text{Area of region }ABFCA:\text{Area of }\triangle BFC$.

Here point F divides the bridge line 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 $\triangle ABD$ with AD as base, applying * area to base division ratio concept*,

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

Similarly in $\triangle ACD$ 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 bridge section point on a median, the bridge line 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 $\triangle BFC$ will be one-third of the main $\triangle ABC$.

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 $\triangle GFP$, $\triangle GEP$ or $\triangle BGC$ are related to each other and to the area of the main $\triangle ABC$.

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

#### Area of $\triangle GFP$ with respec to area of $\triangle GFE$

As median AD is the line from vertex to the last triangle base BC, passing through the bases of triangles, $\triangle AFE$ and $\triangle AGE$, 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 $\triangle GFE$ 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 $\triangle GFE$ with respect to area of $\triangle ABC$

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 $\triangle GFP$ 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 bridge line section 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 $\triangle ABC$ 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 $\triangle ABC$.

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 $\triangle ADQ$ that is made up of medians ot the original $\triangle ABC$.

Thus AR divides the area of $\triangle ADQ$ 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 $\triangle ABD=A_{ABD}=y$,

Area of $\triangle ADR=A_{ADR}=x$, and

Area of $\triangle DCR=A_{DCR}=z$.

And area of $\triangle ABC$ is, $A_{ABC}=2y$, and area of $\triangle ADQ$ 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 $\triangle ABC$ 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 $\triangle ABC$ of side length $a$ is perpendicular to opposite side BC bisecting it, the area of the triangle is,

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

By the use of Pythagorean theorem we have,

$AD=m=\sqrt{a^2-\left(\frac{a}{2}\right)^2}=\sqrt{\frac{3}{4}a^2}=\frac{\sqrt{3}}{2}a$.

So,

$A_{ABC}=\frac{\sqrt{3}}{4}a^2=\frac{1}{\sqrt{3}}m^2$, where $a$ is the side length of the equilateral $\triangle ABC$ 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.

**Related resources that should be useful for you**

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* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

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**section on SSC CGL****generally for any hard MCQ test.**

**Concept tutorials for SSC CGL and other competitive exams on Geometry**

**Basic and rich Geometry concepts part 7, Laws of sines and cosines**

**Basic and rich Geometry concepts part 6, proof of triangle area from medians**

**Basic and rich Geometry concepts part 5, proof of median relations**

**Basic and rich Geometry concepts part 4, proof of arc angle subtending concept**

**Geometry, basic and rich concepts part 3, Circles**

**Geometry, basic concepts part 2, Quadrilaterals polygons and squares**

**Geometry, basic concepts part 1, points lines and triangles**

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