1. Day 67 – Proof of the triangle midpoint theorem

2. Introduction
Geometry deals with geometric figures and their properties. Among the geometrical figures that we have dealt with in details is a triangle. However, we are not done with it. A triangle has other properties other than the basic ones; sides and angles. We also have properties that deal with midpoints of sides and the sides themselves. In this lesson, we will explore the relationship between a line joining the midpoints of any two sides of a triangle and the third side, which is aptly referred to as the triangle midpoint theorem.

3. Vocabulary
1. Midpoint
A point on a line segment that divides it into two equal parts. The point is equidistant from both endpoints of the segment.
2. Theorem
A statement that is can be proved to be true using the already known facts and ideas.

4. The midpoint theorem
The midpoint theorem shows the relationship that exists between the line joining the midpoints of two sides of a triangle and the third side of the triangle.
It states that a line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

5. Consider ∆PQR below. S and T are the midpoints of PQ and PR respectively.

6. According to the triangle midpoint theorem: 𝐒𝐓∥𝐐𝐑 and 𝐒𝐓= 𝟏 𝟐 𝐐𝐑 We can now prove the midpoint theorem.

7. Proof of the midpoint theorem
In order to prove the midpoint theorem, we need to recall the major properties of angles formed when a transversal intersects a pair of parallel lines, especially alternate and corresponding angles. These angles together with vertical angles are always congruent.

8. Consider ∆ABC shown below. We have to show that DE∥BC and DE= 1 2 BC

9. D and E are the midpoints of AB and AC respectively
D and E are the midpoints of AB and AC respectively. From ∆ABC above: AE≅CE and AD≅BD We can extend DE to point F using a broken line segment in such a way that DE≅FE and then draw another broken segment through point C to meet F as shown below. This leads to the formation of ∆CEF. Let us consider ∆ADE and ∆CEF. From the two triangles AE≅CE

10. ∠AED and ∠CEF are vertical angles vertical angles are congruent hence ∠AED≅∠CEF DE≅FE after the extension of DE . A B C D E F

11. ∠AED≅∠CEF ,DE≅FE and AE≅CE Now, using the S. A
∠AED≅∠CEF ,DE≅FE and AE≅CE Now, using the S.A.S postulate, we can conclude that ∆ADE≅∆CEF Corresponding parts of congruent triangles are congruent (CPCTC), therefore: ∠ADE≅∠CFE, ∠DAE≅∠ECF and AD≅CF

12. In the diagram below, ∠ADE≅∠CFE (alternate interior angles) ∠DAE≅∠ECF (alternate interior angles) This shows that AB∥FC implying that DB∥FC and DB≅FC A B C D E F

13. This leads to the formation of parallelogram BCFD and basing on the properties of a parallelogram, we find out that: DF∥BC (opposite sides of a parallelogram are parallel) DF≅BC (opposite sides of a parallelogram are congruent) It then follows that 𝐃𝐄∥𝐁𝐂 Since DE≅EF and DF≅BC, it also follows that DE≅ 1 2 DF, hence 𝐃𝐄≅ 𝟏 𝟐 𝐁𝐂 We have proved the triangle midpoint theorem.

14. The converse of the triangle midpoint theorem is equally important, and it states that if a line segment is drawn from the midpoint of one side of a triangle, parallel to a second side, then the segment bisects the remaining third side.

15. Example In the figure below UQST is a parallelogram in ΔPQR
Example In the figure below UQST is a parallelogram in ΔPQR. QS is produced to point R such that QS≅SR and PT is also produced to point R. Show that PT≅RT by completing the table below. P Q R U S T

16. Statements
Reasons
QU∥ST
Since S is the midpoint of QR, it follows that T is also the midpoint of PR
Hence PT≅RT

17. Solution
Statements
Reasons
QU∥ST
Opposite sides of a parallelogram are parallel
Since S is the midpoint of QR, it follows that T is also the midpoint of PR
Triangle midpoint theorem
Hence PT≅RT

18. homework
D, E and F are the midpoints of the sides on triangle ABC. Show that FBDE is a parallelogram using the triangle midpoint theorem and filling the table below. A B C F D E

19. Statements
Reasons
𝐷𝐸∥𝐵𝐹
FE∥BD
Triangle midpoint theorem
FE≅ 1 2 BC but BD≅ 1 2 BC
DE≅BF
DE≅ 1 2 BA but BF≅ 1 2 BA
Hence FBDE is a parallelogram

20. Statements
Reasons
𝐷𝐸∥𝐵𝐹
Triangle midpoint theorem
FE∥BD
FE≅BC
FE≅ 1 2 BC but BD≅ 1 2 BC
DE≅BF
DE≅ 1 2 BA but BF≅ 1 2 BA
Hence FBDE is a parallelogram

21. THE END