## Knowledge of the mechanism behind a concept increases your belief and effectiveness of its use

Medians form one of the most important set of components in a triangle closely tied to the triangle independent of any other geometric shape. For example another pair of important components, the incentre and the inradius inherit all the properties of a circle to enrich the concept of a triangle, whereas the medians and their intersection point, the centroid, throw more light on the the triangle independent of any other geometric shape.

In this session we will go through the important concepts related to medians and also the mechanisms or proofs behind the concepts.

### What are medians and the centroid

In any discussion on geometry the first thing that we need is a geometric figure. The following figure shows our objects of interest, the medians.

The definition of a median is,

A median is the line segment joining a vertex of a triangle and the mid-point of the opposite side.

In other words, a median is the side bisector passing through a vertex. In the above figure, the medians from three vertices A, B and C are respectively, AD, BE and CF where D, E and F are the midpoints of the sides BC, CA and AB respectively.

The three medians pass through or intersect with each other at a single point inside the triangle G (in our figure). This point of intersection of medians lies inside the triangle and is called the Centroid.

### Concepts on medians and the centroid

The important concepts on medians and centroid are,

- A median is the line segment
This is by definition.**from a vertex of a triangle to the midpoint of the opposite side.** - The
The name of the intersection point is centroid by definition but the single point intersection concept needs proof which we will discuss here.**three medians in a triangle intersect at a single point which is called Centroid.** - The
never outside it. This is a property of centroid that needs proof that we will discuss here.**centroid will always lie inside the the triangle,** - The
Briefly it means, if you consider a triangle as a thin metal plate of absolutely uniform thickness and density, you can balance the triangular metal plate horizontally on a vertically held pencil with the centroid placed on the tip of the pencil, at least theoretically.**centroid is called the centre of gravity of the triangle.** - The
the longer side being towards the vertex. This property needs proof that we will discuss here.**centroid divides each median into two segments of length ratio 2 : 1,** - The
This needs proof and we will show it.**sum of length of two adjacent sides of the triangle is always greater than the length of the median from the common vertex of the two sides.** - The
This follows from the previous property and its proof will follow automatically.**perimeter of the triangle will always be larger than the sum of lengths of the three medians.** We will discuss its proof.**Three times the sum of squares of three sides of the triangle is equal to four times the sum of squares of the length of the medians.**

### Three medians intersect at a single internal point that divides each median in segments in length ratio 2 : 1, mechanism and proof

We will use the following figure for explaining these basic concepts on medians and centroid.

The three concepts we will prove here are,

- Intersection point of medians divide each median in a ratio 2 : 1, with the larger segment lying towards the vertex and smaller towards the base,
- The three medians intersect each other at a single point, the centroid, and
- The centroid always lies inside the triangle.

#### First stage of solution

In $\triangle ABC$, let us consider the two medians AD and CF intersecting each other at G. As F and D are mid-points of two sides of the triangle, the line segment joining these two points FD will always be parallel to the base AC in this case. Consequently the two triangles $\triangle BAC$ and $\triangle BFD$ will be similar to each other. This is a rich concept for the proof of which you may refer to our discussion **Geometry basic and rich concepts part 1, points lines and triangles.**

As these * two triangles are similar, the ratio of corresponding sides are equal to each other.* That is,

$\displaystyle\frac{BF}{BA}=\frac{BD}{BC}=\frac{FD}{AC}=\frac{1}{2}$.

We wanted * primarily the information that FD is half of AC* and we got it.

#### Second stage of solution

We will now focus our attention to the two triangles $\triangle FDG$ and $\triangle ACG$. AS FD || AC the opposite internal angles are equal,

$\angle DFG=\angle ACG$,

$\angle FDG = \angle CAG$,

so that the third angles are also equal,

$\angle DGF = \angle CGA$.

Thus the two triangles are also similar (three angles equal, satisfies AAA similarity test). We have used the property of FD || AC from the first stage of solution. This is use of a more general * geometric property transmission principle. *The parallelism between the two lines FD and AC originated in triangles $\triangle BAC$ and $\triangle BFD$ and is transmitted to the second pair of triangles $\triangle FDG$ and $\triangle ACG$. In geometry frequently use of this principle leads us to the solution quickly.

Because of the similarity of two triangles, $\triangle FDG$ and $\triangle ACG$, the ratio of corresponding sides are equal, that is,

$\displaystyle\frac{GF}{CG}=\frac{GD}{AG}=\frac{FD}{AC}=\frac{1}{2}$.

It means, **each of the two medians AD and CF are divided into two segments of ratio 2 : 1 at their intersection point G with the larger segment lying towards the vertex.**

#### Third stage of solution

In this final stage we will consider now the pair of medians AD and BE, and in the same way can prove that their intersection point, say, $G_1$ divides the two medians in segments of ratio 2 : 1 with larger segment lying towards the vertex. But as both $G$ and $G_1$ are at one third the distance of AD towards A from D, these two points are coincident. This single point of intersection of three medians is named as centroid.

As the centroid lie on a median always between the vertex and the mid-point of the opposite side, it will always lie inside the triangle.

Similar to the concept of centre of a circle as the centre of gravity of the circle, the centroid also can be considered as the centre of gravity of the assymetric shape of a triangle. We will not discuss this point further here and will take up the next concept for showing you the mechanism behind it.

### Sum of two sides is larger than the median from the common vertex, mechanism and proof

This is a rich concept relating sum of lengths of two sides to the length of the common point median. We will use the following figure to show the mchanism behind this rich concept.

In the single order of power unity, the side lengths of a triangle are related to the lengths of medians in two ways that itself are interdependent. Single order means power of length of sides and medians in the relation are all 1.

The * two relations* we will prove now with respect to the above figure are,

- $AB + ACĀ \gt 2AD$, and the same for the other two medians and corresponding pairs of sides. In descriptive form it means, sum of adjacent sides is greater than twice the length of the median from the vertex common to the two sides.
- $AB + BC + CA > AD + BE + CF$. In descriptive terms it means, sum of length of sides is greater than sum of length of medians.

Let us take the first relation to see how this concept happens to be true.

**First stage solution**

In the above figure the $\triangle BPC$ is a mirror image of the $\triangle ABC$ formed by drawing sides BP || AC and CP || AB forming a parallelogram ABPC with two diagonals AP and BC that are bisected at the intersection point D (from parallelogram properties).

AD is the median of interest here.

By the most basic triangle properties, in $\triangle ABP$, sum of any two sides being larger than the third side,

$AB + BP \gt AP$.

But firstly as in a parallelogram opposite sides are equal in length,

$BP = AC$.

And also as the diagonals are inter-bisected at their intersection point,

$DP = AD$,

Or, $AP = 2AD$.

So combining these results we have,

$AB + AC \gt 2AD$.

**Second stage solution**

Similarly,

$BC + BA \gt 2BE$, and,

$CB + CA \gt 2CF$.

Summing the three inequations,

$2(AB + BC + CA) \gt 2(AD + BE + CF)$,

Or, $AB + BC + CA \gt AD + BE + CF$.

These are then the mechanisms behind the two first order relations between triangle sides and medians.

### Relation between sums of squares of medians and squares of sides, proof

The medians and sides of a triangle are related in a second order equation of squares that is a surprise and a rich concept a bit advanced. Nevertheless we will try to keep the proof of this important rich concept as simple as possible dependent only on the most basic geometric concepts.

In descriptive terms the second order relation between the sides and medians of a triangles is,

Three times the sum of squares of the length of sides equals four times the squares of medians of a triangle.

We will show how this happens to be by using the following figure,

By the second order relation with respect to the above figure where D, E and F are the mid-points of the three sides BC, CA and AB of $\triangle ABC$ and the medians are AD, BE and CF,

$3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$

This is also known as Apollonius' theorem.

Similar to the linear relations between medians and sides, here also two relations are involved, the second one following from the first. The two second order interdependent relations are,

- $AB^2 + AC^2 = 2AD^2 + \displaystyle\frac{BC^2}{2}$, and
- $3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$.

#### First stage solution

In $\triangle ABC$, AH is the perpendicular from A to D and is the height of the triangle in this configuration whereas G is the centroid.

For convenience, let us assume, $AB=c$, $BC=a$, $CA=b$, $AH=h$, $DH=x$ and median $AD=m_1$.

Here as AD is the first median and it bisects BC into two equal parts, $BD=DC=\displaystyle\frac{a}{2}$.

In right $\triangle AHC$,

$h^2 = b^2 - HC^2 $

$\hspace{7mm}= b^2 - (\frac{a}{2} - x)^2$

$\hspace{7mm}=b^2 - (\frac{a^2}{4} -ax + x^2)$

Or, $b^2 + ax - \frac{a^2}{4} = h^2 + x^2$

But in right $\triangle AHD$,

$h^2 + x^2= m_1^2$.

So,

$b^2 + ax - \frac{a^2}{4} = m_1^2$.

Similarly in $\triangle AHB$,

$h^2 = c^2 - HB^2 $

$\hspace{7mm}= c^2 - (\frac{a}{2} + x)^2$

$\hspace{7mm}=c^2 - (\frac{a^2}{4} +ax + x^2)$,

Or, $c^2 - ax - \frac{a^2}{4} = h^2 + x^2 = m_1^2$.

Adding the two,

$b^2 + c^2 = 2m_1^2 + \frac{a^2}{2}$.

Or, $AB^2 + AC^2 = 2AD^2 + \displaystyle\frac{BC^2}{2}$.

#### Second stage solution

In the same way,

$c^2 + a^2 = 2m_2^2 + \frac{b^2}{2}$, and

$a^2 + b^2 = 2m_3^2 + \frac{c^2}{2}$.

Adding the three equations we get,

$2(a^2 + b^2 + c^2) = 2(m_1^2 + m_2^2 + m_3^2) + \frac{1}{2}(a^2+b^2+c^2)$

Or, $\frac{3}{2}(a^2 + b^2+c^2) = 2(m_1^2 + m_2^2 + m_3^2)$,

Or, $3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$.

The solution is based on very basic concepts and also the process is simple enough to be understood and believed by anyone. Belief on a new concept is vital to usable knowledge.

You may refer to other proofs of Apollonius' theorem to have a comparative understanding of various approaches to a problem.

We will end this discussion with a problem at the SSC CGL exam level.

### Problem on median side relation

In $\triangle ABC$ with medians AD, BE and CF, (D, E and F being the midpoints of the sides BC, CA and AB respectively) the following is true,

- $4(AB^2 + BC^2 + CA^2) < 3(AD^2 + BE^2 + CF^2)$
- $AB^2 + BC^2 + CA^2 > \displaystyle\frac{1}{2}(AD^2 + BE^2 + CF^2)$
- $3(AB^2 + BC^2 + CA^2) > 4(AD^2 + BE^2 + CF^2)$
- $(AB^2 + BC^2 + CA^2) > 2(AD^2 + BE^2 + CF^2)$

#### Problem analysis and Solution.

The figure depicting the problem is shown below.

The * rich concept of median to side length relation* in a triangle states that,

Three times the sum of squares of three sides of a triangle equals four times the sum of squares of its medians.

To state the concept in terms of algebraic relation, with respect to the figure above the following relation between the sum of sides and sum of medians holds good,

$3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)$.

Or, $(AB^2 + BC^2 + CA^2) = \frac{4}{3}(AD^2 + BE^2 + CF^2)$,

Or, $(AB^2 + BC^2 + CA^2) \gt \frac{1}{2}(AD^2 + BE^2 + CF^2)$, as $\frac{4}{3} \gt \frac{1}{2}$.

All other options are tested to be invalid.

**Answer:** Option b: $(AB^2 + BC^2 + CA^2) \gt \frac{1}{2}(AD^2 + BE^2 + CF^2)$.

**Key concepts used:** Rich concept of * Median to side length second order relation *--

**inequality analysis.****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**

**You may refer to:**

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

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**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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