In geometry, the mid-point theorem helps us to find the missing values of the sides of the triangles. It establishes a relation between the sides of a triangle and the line segment drawn from the midpoints of any two sides of the triangle. The midpoint theorem states that the line segment drawn from the midpoint of any side to the midpoint of any other side of a triangle is parallel to the third side and is half of the length of the third side of the triangle.
In this article, we will explore the concept of the midpoint theorem and its converse. We will learn the application of the theorem with the help of a few solved examples for a better understanding of the concept.
| 1. | What is Midpoint Theorem? |
| 2. | State and Prove Mid Point Theorem |
| 3. | Converse of MidPoint Theorem |
| 4. | Application of Midpoint Theorem |
| 5. | FAQs on Midpoint Theorem |
The midpoint theorem states that "the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of the length of the third side". It is often used in the proofs of congruence of triangles.
Consider an arbitrary triangle, ΔABC. Let D and E be the midpoints of AB and AC respectively. Suppose that you join D to E. The midpoint theorem says that DE will be parallel to BC and equal to exactly half of BC. Look at the image given below to understand the triangle midpoint theorem.
Statement: The midpoint theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of the third side.
i.e., in a ΔABC, if D and E are the midpoints of AB and AC respectively, then DE || BC and DE = ½ BC.
Now, let us prove the midpoint theorem. Consider the triangle ABC, as shown in the figure below. Let E and D be the midpoints of the sides AC and AB respectively.
Given: D and E are the mid-points of sides AB and AC of ΔABC respectively.
To Prove: DE || BC and DE = 1/2 × BC
Construction: In ΔABC, through C, draw a line parallel to BA, and extend DE such that it meets this parallel line at F, as shown below:
| Mid Point Theorem Proof | |
|---|---|
| Statement | Reason |
| Compare ΔAED with ΔCEF: | |
| 1. AE = EC | E is the midpoint of AC (Given) |
| 2. ∠DAE = ∠FCE | alternate interior angles |
| 3. ∠DEA = ∠FEC | vertically opposite angles |
| 4. ΔAED ≅ ΔCEF | By the ASA criterion |
| 5. DE = EF and AD = CF | By CPCTC |
| 6. AD = BD | D is the mid point of AB (Given) |
| 7. BD = CF | From 5 and 6 |
| 8. BCFD is a parallelogram. | BD || CF (by construction) and BD = CF (from 7) |
| 9. DF || BC and DF = BC | BCFD is a parallelogram |
| 10. DE || BC | DE is part of DF and from 9 |
| 11. DE + EF = BC | E is a point on the line segment DF |
| 12. 2DE = BC | DE = EF from 5 |
| 13. DE = 1/2 × BC | Dividing both sides by 2 |
| Midpoint theorem is proved by 10 and 13 | |
Will the converse of the midpoint theorem hold? Yes, it will, and the proof of the converse is presented next.
Statement: The converse of midpoint theorem states that "the line drawn through the midpoint of one side of a triangle that is parallel to another side will bisect the third side". We prove the converse of mid point theorem by contradiction.
Consider a triangle ABC, and let D be the midpoint of AB. A line through D parallel to BC meets AC at E, as shown below.
Given: In ΔABC, D is the midpoint of AB and DE || BC.
To Prove: E is the midpoint of AC (i.e., AE = CE)
Construction: Through C, draw a line parallel to AB that meets the extended DE at F.
| Proof of Converse of Midpoint Theorem | |
|---|---|
| 1. BCFD is a parallelogram | DE || BC (given) and BD || CF (by construction) |
| 2. BD = CF | Opposite sides of a parallelogram are equal |
| 3. AD = BD | D is the midpoint of AB (given) |
| 4. AD = CF | from 2 and 3 |
| Compare ΔAED with ΔCEF: | |
| 5. ∠DAE = ∠ECF | Alternative angles |
| 6. ∠DEA = ∠FEC | Vertically opposite angles |
| 7. ΔAED ≅ ΔCEF | By AAS criterion (using 4, 5, and 6) |
| 8. AE = CE | By CPCTC |
This completes our proof of the converse midpoint theorem.
An interesting consequence of the midpoint theorem is that if we join the midpoints of the three sides of any triangle, we will get four (smaller) congruent triangles, as shown in the figure below:
We have: ΔADE ≅ ΔFED ≅ ΔBDF ≅ ΔEFC.
Proof: Consider the quadrilateral DEFB. By the midpoint theorem, we have:
Thus, DEFB is a parallelogram, which means that ΔFED ≅ ΔBDF. Similarly, we can show that AEFD and DECF are parallelograms, and hence all four triangles so formed are congruent to each other (make sure that when you write the congruence relation between these triangles, you get the order of the vertices correctly).
☛ Related Topics:
Important Notes on Midpoint Theorem:
Example 1: Consider a triangle ABC, and let D be any point on BC. Let X and Y be the midpoints of AB and AC respectively. Show that XY bisects AD. Solution: It is given that X and Y are the midpoints of AB and AC. By the midpoint theorem, XY || BC. Now, consider ΔABD. The segment XE is parallel to the base BD, and X is the midpoint of AB. By the converse of the midpoint theorem, E must be the midpoint of AD. Thus, XY bisects AD. Answer: It is proved that XY bisects AD at E.
Example 2: Prove that if three parallel lines make equal intercepts on one transversal, then they will make equal intercepts on any other transversal as well. Solution: Let us first understand this problem in a better way. Consider three lines and two transversals, as shown below: 
Suppose that the intercepts on the left transversal are equal, that is, AB = BC. We then have to prove that the intercepts on the right transversal will also be equal, that is, DE = EF. To prove this, join A to F: 
Consider ΔACF. Since B is the midpoint of AC and BG || CF, the converse of the midpoint theorem tells us that G is the midpoint of AF. Now, consider ΔAFD. We have shown that G is the midpoint of AF. Also, GE || AD. Thus, the converse of the midpoint theorem tells us that E must be the midpoint of FD. Therefore, DE = EF. Answer: It is proved that DE = EF.
Example 3: 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. Solution: Consider the following figure: We have to show that BX = XY = YD = BD/3. First of all, we note that AECF is a parallelogram as ABCD is a parallelogram which means AB = CD and so AE = CF (as E and F are the mid-points), and thus, EC || AF. Now, consider ΔBAY. Since E is the midpoint of AB, and EX || 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. Thus, BX = XY = YD = BD/3. Answer: Hence proved.
