Medians of a Triangle: A triangle is a three-sided polygon having three sides, three angles, and three vertices. It is one of the most fundamental geometric forms. A triangle’s median is the line segment that connects a triangle’s vertex to the middle of the opposing side, thereby bisecting that side. There are three medians for each triangle, one from each vertex.

These medians cross at a point, which is known as the centroid of a triangle and is represented by the letter G. The centroid of a triangle is the place at which its medians connect, and it (centroid) divides each median in the ratio of 2:1. Embibe provides the guidelines from subject experts for the students to go through the topics thoroughly. This article will go through medians, their characteristics, and solved instances.

Medians of a Triangle: Definition

The median of a triangle is the line segment joining a vertex to the mid-point of the other side of a triangle. A triangle contains three medians, one from each vertex. The three medians are concurrent at a point called the centroid of the triangle.

Here, \(DM,\,EN\) and \(LF\) are medians intersect each other at the centroid \(G\).

Centroid divides the median in the ratio \(2:1.\)

Altitude: The altitude of a triangle is perpendicular from a vertex to the opposite side of a triangle is called an altitude. A triangle contains three altitudes, one from each vertex. The three altitudes are concurrent at a point called the orthocentre of the triangle.

Here, \(DL\) is an altitude from the vertex \(D\) to its opposite side \(EF\).

Medians of a Triangle: Concurrent

The point at which unparallel straight lines intersect or meet is called the point of concurrence.

The medians of a triangle are concurrent. The point of concurrency of medians is called the centroid of the triangle. The medians of a triangle are concurrent in the internal part of a triangle. The centroid splits the medians into a \(2:1\) ratio.

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Proof of medians of a triangle are concurrent:

In the first figure above, \(QE\) and \(DR\) are median on \(PR\) and \(PQ\) respectively. So, \(D\) and \(E\) are midpoints on \(PQ\) and \(PR\), respectively.

Therefore, \(DE\,{\rm{||}}\,QR\) and \(DE\, = \,\frac{1}{2}\,QR\) [Midpoint theorem]
\(\frac{{DE}}{{QR}}\, = \,\frac{1}{2} –  –  –  –  – ({\rm{I}})\)

Now, consider \(\Delta \,DEG\) and \(\Delta \,QGR\)
\(\angle GQR = \,\angle GED\, = \,y\) and \(\angle RDE = \,\angle DRQ\, = \,x\)

Therefore, from \(AA\) similarity criterion \(\Delta DEG \sim \Delta QGR.\)

Then, \(\frac{{EG}}{{GQ}} = \frac{{DG}}{{GR}} = \frac{{DE}}{{QR}} –  –  –  – ({\rm{II}})\)

From equation \(({\rm{I}})\) and \(({\rm{II}})\)

\(\frac{{EG}}{{GQ}} = \frac{{DG}}{{GR}} = \frac{{DE}}{{QR}} = \frac{1}{2} –  –  –  – ({\rm{III}})\)

Similarly, from the second figure, we get \(\Delta PQG \sim \Delta FEG.\)

\(\frac{{EG}}{{GQ}}\, = \,\frac{{EF}}{{PQ}}\, = \,\frac{{FG}}{{GP}}\, = \,\frac{1}{2} –  –  –  – ({\rm{IV}})\)

From equation \(({\rm{III}})\) and \(({\rm{IV}})\)

\(\frac{{EG}}{{GQ}}\, = \,\frac{{DG}}{{GR}}\, = \,\frac{{FG}}{{GP}}\, = \,\frac{1}{2}\)

Hence, proved.

Medians of a Triangle: Formula

Let us have a look at how to calculate the length of each median:

\({m_a}\, = \,\sqrt {\frac{{2{b^2} + 2{c^2} – {a^2}}}{4}} \)
\({m_b} = \,\sqrt {\frac{{2{a^2} + 2{c^2} – {b^2}}}{4}} \)
\({m_c} = \sqrt {\frac{{2{a^2} + 2{b^2} – {c^2}}}{4}} \)
\(a,b\) and \(c\) are three edges of the triangle.

Centroid of a Triangle Formula

The centroid of a triangle takes the average of the \(x\) coordinates and the \(y\) coordinates of all the three vertices. Thus, the centroid formula can be mathematically stated as \(G(x,y) = \left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)

Medians of a Triangle Properties

Properties of medians of a triangle are as follows:

1. The median of a triangle further divides the triangle into two triangles having the exact area measurement. 
2. For a given triangle, the second median divides the triangle formed by the first median in the ratio \(1:2.\)
3. Each triangle has three medians, one from every vertex. The point of concurrency of three medians forms the centroid of the triangle.
4. Regardless of the shape or size of a triangle, its three medians meet at a single point. 
5. Each median of a triangle divides the triangle into two smaller triangles that have equal areas. The three medians divide the triangle into \(6\) smaller triangles of similar area.
6. The sum of two sides of a triangle is greater than the median drawn from the vertex, which is expected.

Medians of an Equilateral Triangle are Equal

The length of medians in an equilateral triangle is always equal. Since the lengths of all sides in an equilateral triangle are the same, the length of medians bisecting these sides are equal.

Here, the medians \(AE,\,BD\) and \(CF\) are equal

Medians of a Triangle Notes

1. The median of any triangle bisects it into two triangles of equal areas.
2. The centroid (the point where they meet) is the centre of gravity of the triangle.
3. Sum of medians of a triangle: The sum of squares of the medians of a triangle equals three-fourths of the sum of squares of the sides of the triangle.
4. The boundary of a triangle is greater than the sum of their three medians.
5. The equivalent sides, boundaries, medians, and heights will all be in the same ratio for two similar triangles. 
6. The midpoint of a segment divides it into two congruent parts, whereas the median of a triangle divides it into two congruent triangles.
7. Also, if two triangles are congruent, the medians of congruent triangles are equal as corresponding parts of congruent triangles are congruent.
8. The centroid of any triangle is a tri-sector point of all the medians.
9. The centroid of any type of triangle always lies inside the triangle.

Solved Examples – Medians of a Triangle

Q.1. \(PM\) is a median of \(\Delta PQR\). \(PM\) is extended to \(N\) such that \(PM = MN\) If \(O\) be the mid- point of \(MN\) and if the area of \(\Delta PQR\) is \(40\,{\rm{sq}}{\rm{.cm}}\) , then find the area of the \(\Delta QOM\).

Ans: \(PM\) is the median of \(\Delta PQR\).
\(\therefore \,\,\left( {\Delta PQM} \right) = \frac{1}{2}ar\left( {\Delta PQR} \right) = \frac{1}{2} \times 40\,{\rm{sq}}{\rm{.cm}}\,{\rm{ = }}\,{\rm{20}}\,{\rm{sq}}{\rm{.cm}}\)
Again, \(QM\) is the median of \(\Delta PQN\).
\(\therefore \,\,ar\left( {\Delta QMN} \right) = \frac{1}{2}ar\left( {\Delta PQN} \right) = \frac{1}{2} \times 40\,{\rm{sq}}{\rm{.cm}}\,{\rm{ = }}\,{\rm{20}}\,{\rm{sq}}{\rm{.cm}}\)
Now \(QO\) is a median of the \(\Delta QMN\).
\(ar\left( {\Delta QOM} \right) = \frac{1}{2}ar\left( {\Delta QMN} \right) = \frac{1}{2} \times 20\,{\rm{sq}}.{\rm{cm}}\,{\rm{ = }}\,{\rm{10}}\,{\rm{sq}}{\rm{.}}\,{\rm{cm}}\)
Therefore, the required area is \({\rm{10}}\,{\rm{sq}}{\rm{.}}\,{\rm{cm}}\).

Q.2. \(E\) is the mid-point of a median \(AD\) of the \(\Delta ABC\). Prove that \(\Delta BED = \frac{1}{4}\Delta ABC\)
Ans:
Proof

\(AD\) is the median of \(\Delta ABC\)
\(\therefore \Delta ABD = \frac{1}{2}\Delta ABC –  –  –  –  – ({\rm{I}})\)
( \(\therefore \) median of every triangle bisects it into the two triangles of equivalent areas.)
Again, \(E\) is the mid-point of \(AD.\)
\(\therefore BE\) is a median of \(\Delta ABD\).
\(\therefore \Delta BED = \frac{1}{2}\Delta ABD = \frac{1}{2} \times \frac{1}{2}\Delta ABC\) (by \(({\rm{I}})\))
\( = \,\frac{1}{4}\Delta ABC\)
\(\therefore \,\Delta BED\, = \,\frac{1}{4}\Delta ABC\)

Q.3. For the given \(\Delta PQR,\,G\) is the centroid and \(QR\, = \,12\) \({\rm{units}}\). Determine the length of \(MR\).

Here \(PM,\,QN,\) and \(RL\) are the medians of \(\Delta PQR,\,M\) is the midpoint of \(QR\) as \(PM\) is the median \(QR\, = 12\,{\rm{units}},\,MR = 6\,{\rm{units}}\). Therefore, \(MR\,\, = \,6\,{\rm{units}}\).

Q.4. Compute the coordinates of the centroid of a triangle for which the vertices are \((2,\,6),\,(8,\,12)\,and\,(8,\,0)\).
Ans: We know that the coordinates of the centroid of a triangle whose angular points are
\(({x_1},\,{y_1}),\,({x_2},\,{y_2}),\,({x_3},\,{y_3})\) are \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\,\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)
Therefore, the coordinates of the centroid of a triangle for which the vertices are \((2,\,6),\,(8,\,12)\) and \((8,\,0)\) are \(\left( {\frac{{2 + 8 + 8}}{3},\,\frac{{6 + 12 + 0}}{3}} \right) = (6,\,6)\)
Hence, the centroid of a triangle for the given vertices is \((6,\,6)\).

Q.5. Prove that if any two medians of a triangle are equal, then it is an isosceles triangle.
Ans: Given: Let the medians \(BE\) and \(CF\) of the \(\Delta ABC\) are equal, i.e., \(BE\, = \,CF\)

To prove : \(\Delta ABC\) is an isosceles triangle.
Proof: Let \(BE\) and \(CF\) intersect each other at \(G.\) We know that the three medians of a triangle intersect at a tri-sector point within it, \(\therefore EG\, = \,\frac{1}{3}\,BE\) and \(FG\, = \,\frac{1}{3}CF\)
But \(BE\, = \,CF\,,\,\therefore \,EG\, = \,FG\, –  –  –  – ({\rm{I}})\) and \(BG\, = \,CG\, –  –  –  – ({\rm{II}})\)
Now, in \(\Delta BGF\) and \(\Delta CGE\), \(FG\, = \,EG\) ( by \(({\rm{I}})\)), \(BG\, = \,CG\) (by \(\left( {{\rm{II}}} \right)\))
And \(\angle BGF\, = \,\angle CGE\) (\(\therefore\) these are opposite angles)
\(\therefore \,\Delta BGF\, \cong \,\Delta \,CGE\) (by the condition of \(SAS\) congruence)
\(\therefore BF = CE\) (\(\therefore\) these are similar sides of congruent triangles)
or, \(\frac{1}{2}AB = \frac{1}{2}AC\) (\(\therefore\) \(F\) and \(E\) are mid-points of \(AB\) and \(AC\) respectively) or, \(AB\, = \,AC\)
\(\therefore \,\Delta ABC\) is an isosceles triangle.
Hence, proved.

Summary

In this article, we learnt about medians of a triangle definition, medians of a triangle are concurrent, medians of a triangle formula, medians of triangle properties, medians of a triangle are equal, medians of a triangle notes solved examples on medians of a triangle, and FAQs on medians of a triangle.

This article’s learning outcome is that we learned the relation between triangle medians and their sides and the connection between a triangle’s area to its medians.

Learn About Properties of a Triangle

Frequently Asked Questions (FAQs) On Medians of a Triangle

Q.1: What are the properties of the median of a triangle?
Ans: Properties of medians of a triangle are as follows:
1. The median of a triangle further divides the triangle into two triangles having the exact area measurement. 
2. For a particular triangle, the second median splits the triangle created by the first median in the ratio \(1:2\)
3. Each triangle has three medians, one from every vertex. The point of concurrency of three medians forms the centroid of the triangle.
4. Regardless of the shape or size of a triangle, its three medians have been met at a single point. 
5. Each median of a triangle divides the triangle into two smaller triangles that have equal areas. The three medians divide the triangle into \(6\) smaller triangles of similar area.
6. The sum of two sides of a triangle is greater than the median drawn from the vertex, which is expected.

Q.2: Are the medians of a triangle equal?
Ans: The length of medians in an equilateral triangle is always equal. Since the length of all sides in an equilateral triangle is equal, it follows that the length of medians bisecting these sides is equal.

Q.3: How do you find the length of a median in a triangle?
Ans: Formula to find the length of each median is
\({m_a} = \sqrt {\frac{{2{b^2} + 2{c^2} – {a^2}}}{4}} \)
\({m_b} = \sqrt {\frac{{2{a^2} + 2{c^2} – {b^2}}}{4}} \)
\({m_c} = \sqrt {\frac{{2{a^2} + 2{b^2} – {c^2}}}{4}} \)
\(a,\,b\) and \(c\) are three edges of the triangle.

Q.4: What does the median mean in a triangle?
Ans: Medians of a triangle is the line segment joining a vertex to the mid-point of the opposite side of a triangle is called a median. A triangle contains three medians, one from each vertex.

Q.5: What is the centroid of a triangle?
Ans: The medians of a triangle are concurrent (they intersect in one common point). The point of the concurrency of medians is referred to as the centroid of the triangle.
The centroid of a triangle holds the mean of the \(x\) coordinates and the \(y\) coordinates of all the three vertices. Thus, the centroid formula can be mathematically stated as \(G(x,\,y)\, = \,\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\)

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