# Midpoint Theorem: Definition, Proofs, Application

Midpoint Theorem states that “the line segment in a triangle crossing the midpoints of two sides of the triangle is said to be parallel to its third side and is also half the length of the third side. In midpoint theorem-proof, we use some geometric properties such as congruence of triangles, pair of angles theorem, parallel lines, etc.

In this article, let us learn the midpoint theorem-proof, its application with examples

## Midpoint Theorem Definition

The mid-point theorem states that if the mid-points of any two sides of a triangle are joined by a line segment, then this line segment is parallel to the third side of the triangle and is half the length of the third side.

That is, in $$\Delta ABC,$$ if $$D$$ and $$E$$ are mid-points of the sides $$AB$$ and $$AC,$$ respectively, then $$DE\parallel BC$$ and $$DE = \frac{1}{2}BC.$$

### Midpoint TheoremProof

Let us consider the $$\Delta ABC.$$
$$E$$ and $$D$$ are the midpoints of the sides $$AC$$ and $$AB.$$
We need to prove that $$DE\parallel BC$$ and $$DE = \frac{1}{2}BC.$$
Now consider the below figure.

Construction: Let us join $$E$$ and $$D$$ and extend the line segment $$DE$$ and produce it to $$F$$ such that, $$EF = DE.$$
Join $$CF.$$
In $$\Delta ADE$$ and $$\Delta CFE,$$
$$EC = AE$$ ($$\because E$$ is the midpoint of the side$$AC$$)
$$\angle CEF = \angle AED$$ (These are vertically opposite angles as $$AC$$ and $$DF$$ intersects at point $$E$$ and vertically opposite angles are equal)
$$EF = DE$$ (By the construction)

By $$SAS$$ congruence criterion,
$$\Delta CFE \cong \Delta ADE$$
We know that the corresponding parts of congruent triangles are equal.
Therefore,
$$\angle CFE = \angle ADE$$ (by c.p.c.t.)
$$\angle FCE = \angle DAE$$ (by c.p.c.t.)
and $$CF = AD$$ (by c.p.c.t.)
$$\angle CFE$$ and $$\angle ADE$$ are the alternate interior angles as $$DA\parallel CF$$ and $$DF$$ is the transversal.

Assume $$CF$$ and $$AB$$ as two lines that are intersected by the transversal $$DF.$$
Similarly, $$\angle FCE$$ and $$\angle DAE$$ are the alternate interior angles.
Assume $$CF$$ and $$AB$$ are the two lines that are intersected by the transversal $$AC.$$
Therefore, $$CF\parallel AB.$$

So, $$CF\parallel BD.$$
and $$CF = BD$$ (since $$BD = AD,$$ it is proved that $$CF = AD$$)
Thus, $$BDFC$$ forms a parallelogram.
We know that opposite sides of a parallelogram are parallel to each other and equal. By the properties of a parallelogram, we can now write:
$$BC\parallel DF$$ and $$BC = DF$$
$$BC\parallel DE$$ and $$DE = \frac{1}{2}BC$$
Hence, the midpoint theorem is proved.

### Converse of Midpoint Theorem

The converse of midpoint theorem states that if a straight line is drawn starting from the midpoint of any side of a triangle and parallel to another side, then it intersects the third side at its mid-point.

In $$\Delta ABC,$$ if the line $$DE$$ is drawn from the midpoint $$D$$ of side $$AB,$$ parallel to $$BC,$$ it intersects the side third side $$AC,$$ such that $$E$$ is the midpoint of $$AC$$ or $$AE = EC.$$

### Solved Examples

Q.1. In $$\Delta ABC,$$ the midpoints of the sides $$BC,\,CA,\,AB$$ are $$D,\,E$$ and $$F,$$ respectively. Find the value of $$EF,$$ if the value of $$BC = 16\,{\rm{cm}}.$$

Ans: Given that the length of $$BC = 16\,{\rm{cm}}$$
Also given that $$F,\,E$$ are the midpoints of $$AB$$ and $$AC.$$
Let us construct a line segment $$FE.$$
The midpoint theorem states that if the midpoints of any two sides of a triangle are joined by a line segment, then this line segment is parallel to the third side of the triangle and is half the length of the third side.
$$\Rightarrow EF = \frac{1}{2}BC$$
Now, substituting the value of $$BC$$ we get $$EF = \frac{1}{2} \times 16\,{\rm{cm = 8}}\,{\rm{cm}}$$
Therefore, the value of $$EF = {\rm{8}}\,{\rm{cm}}{\rm{.}}$$

Q.2. In $$\Delta ABC,\,D,\,E$$ and $$F$$ are respectively the mid-points of sides $$AB,\,BC$$ and $$CA.$$ Show that $$\Delta ABC$$ is divided into four congruent triangles by joining $$D,\,E$$ and $$F.$$

Practice Exam Questions

Ans: It is given that in $$\Delta ABC,\,D,\,E$$ and $$F$$ are the mid-points of the sides $$AB,\,BC$$ and $$CA$$ respectively.
According to the mid-point theorem, $$DE\parallel AC$$
Similarly, $$DF\parallel BC$$ and $$EF\parallel AB.$$
Therefore, $$ADEF,\,BDFE$$ and $$DFCE$$ are all parallelograms.
Now, $$DE$$ is a diagonal of the parallelogram $$BDFE,$$
Thus, $$\Delta BDE \cong \Delta FED$$ (as a diagonal of a parallelogram divides it into two congruent triangles)

Similarly, $$FE,$$ is a diagonal of the parallelogram $$DFCE$$
Thus, $$\Delta FED \cong \Delta CFE$$
Similarly, $$DF$$ is a diagonal of the parallelogram $$ADEF$$
$$\Delta FED \cong \Delta ADF$$
Hence, $$\Delta ABC$$ can be divided into four congruent triangles by joining $$D,\,E$$ and $$F.$$

Q.3. Consider a $$\Delta ABC,$$ and let $$D$$ be any point on $$BC.$$ Let $$X$$ and $$Y$$ be the midpoints of $$AB$$ and $$AC.$$ Show that $$XY$$ will bisect $$AD.$$

Ans: Given that $$X$$ and $$Y$$ are the midpoints of $$AB$$ and $$AC$$ in $$\Delta ABC.$$
The midpoint theorem states, that if the mid-points of any two sides of a triangle are joined by a line segment, then this line segment is parallel to the third side of the triangle and is half the length of the third side.
Thus, $$XY$$ parallel $$BC.$$
Now, consider $$\Delta ABD$$
The line segment $$XE$$ is parallel to the base $$BD\left( {XY\parallel BC} \right)$$
Now, according to the converse midpoint theorem if a straight line is drawn starting from the midpoint of any side of a triangle and parallel to another side, then it intersects the third side at its mid-point.
Thus, $$X$$ is the midpoint of $$AB$$ and $$E$$ must be the midpoint of $$AD.$$
Hence, $$XY$$ bisects $$AD.$$

Q.4. Prove that if three parallel lines make equal intercepts on one transversal, then they will make equal intercepts on any other transversal as well.
Ans: Let us look into understanding the problem better. Consider these three lines and two transversals, as shown in the image below:

Suppose that the intercepts on the left transversal are equal, that is, $$AB = BC.$$
We have to prove that the intercepts on the right transversal will also be equal, that is, $$DE = EF.$$
Construction: Join $$A$$ to $$F.$$

Let us consider $$\Delta ACF.$$
Since $$B$$ is the midpoint of $$AC$$ and $$BG\parallel CF,$$ as $$BE\parallel CF$$ the converse of the midpoint theorem tells us that $$G$$ is the midpoint of $$AF.$$
Now, consider $$\Delta AFD$$ we have shown that $$G$$ is the midpoint of $$AF.$$ Also, $$GE\parallel AD.$$
Thus, the (converse of the) midpoint theorem tells us that $$E$$ must be the midpoint of $$FD.$$
Hence, $$DE = EF.$$

Q.5. Consider a parallelogram $$ABCD.\,E$$ and $$F$$ are the midpoints of $$AB$$ and $$CD$$ respectively. Show that the line segments $$AF$$ and $$EC$$ trisect the diagonal $$BD.$$
Ans:
Let us consider a parallelogram $$ABCD,$$ and $$E,\,F$$ are the midpoints of $$AB$$ and $$CD$$ respectively.
To prove: The line segments $$AF$$ and $$EC$$ trisects the diagonal $$BD.$$
$$\left[ {BX = XY = YD = \frac{1}{3}BD} \right]$$
Construction: Join $$A,\,F$$ and $$E,\,C.$$

We can observe that $$AECF$$ is a parallelogram ($$AE\parallel CF$$ and $$AE = CF$$)
Thus, $$EC\parallel AF.$$
Now, consider $$\Delta AYB$$

Since $$E$$ is the midpoint of $$AB,$$ and $$EX\parallel AY,$$ the (converse of the) midpoint theorem tells us that $$X$$ is the midpoint of $$BY,$$ which means that $$BX = XY.$$
Similarly, we can prove that $$XY = YD.$$
Hence, $$BX = XY = YD = \frac{1}{3}BD$$ (proved)

### Summary

We have learned in this article about the midpoint theorem and its application. We have discussed the converse of mid-point theorem in brief. We have learned how these theorems help us to prove other geometric proofs such as if we join the midpoint of any quadrilateral in order we get a parallelogram, intercept theorem, etc.

### FAQs

Q.1. What is the midpoint theorem?
Ans: Midpoint theorem states, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of the length of the third side. Using this, we can find the missing sides of the triangle.

Q.2. How to find the midpoint in coordinate geometry if the coordinates of endpoints of a line segment given?
Ans:
A midpoint means a point that lies at or near the middle or equidistant from both ends of the line segment. In other words, the midpoint is the position of a fact which is halfway between the beginning and the end.
If $$P\left( {{x_1},\,{y_1}} \right)$$ and $$Q\left( {{x_2},\,{y_2}} \right)$$ are the coordinates of two given endpoints, then the midpoint formula is as follows:
Midpoint $$= \left[ {\frac{{{x_1} + {x_2}}}{2},\,\frac{{{y_1} + {y_2}}}{2}} \right]$$

Q.3. What is the converse of midpoint theorem?
Ans: Converse midpoint theorem states that if a straight line is drawn starting from the midpoint of any side of a triangle and parallel to another side, then it intersects the third side at its mid-point.

Q.4. What do we understand by a midpoint in geometry?
Ans: In geometry, the midpoint is defined as the point which divides the line segment into two equal parts. The midpoint is equidistant from the endpoints of a line segment.

Q.5. What is the midpoint theorem in a quadrilateral?
Ans:
We can apply the midpoint theorem to prove other geometric properties. We can apply it in the quadrilateral as well. If we join the midpoints of each side, it gives a parallelogram. To prove it, we need to construct one of the diagonals of the quadrilateral that we can apply the midpoint theorem of a triangle.

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