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. Consider an arbitrary triangle,
Δ
A
B
C
. Let D and E be the midpoints of AB and AC. Suppose that you join D and E:
Arbitrary triangle
The Midpoint Theorem says that DE will be parallel to BC and equal to exactly half of BC.
Visualise the Midpoint Theorem
In the simulation below, drag the vertices of the triangle to see the Midpoint Theorem in action. DE is always parallel to BC, and equal to exactly half of BC:
How do you Prove the Midpoint Theorem?
Proof: Through C, draw a line parallel to BA, and extend DE such that it meets this parallel at F, as shown below:
Midpoint Theorem
Compare
Δ
A
E
D
with
Δ
C
E
F
:
1. AE = EC (E is the midpoint of AC)
2.
∠
D
A
E
=
∠
F
C
E
(alternate interior angles)
3.
∠
D
E
A
=
∠
F
E
C
(vertically opposite angles)
By the ASA criterion, the two triangles are congruent. Thus, DE = EF and AD = CF. But AD is also equal to BD, which means that BD = CF (also, BD || CF by our construction). This implies that BCFD is a parallelogram. Thus,
1. DF || BC è DE || BC
2. DE = EF = ½(DF) = ½(BC) èDE = ½(BC)
This completes our proof. Will the converse of this theorem hold? Yes, it will, and the proof of the converse is presented next.
What is Converse Midpoint Theorem?
The Converse Midpoint Theorem states that the line drawn through the midpoint of one side of a triangle which is parallel to another side will bisect the third side. 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:
Converse of the Midpoint Theorem
The Converse of the Midpoint Theorem says that E is the midpoint of AC.
Proof of the Converse of the Midpoint Theorem
Proof: Suppose that E is not the midpoint of AC. Let F be the midpoint of AC. Join D to F, as shown below:
Midpoint Theorem Exercise
By the Midpoint Theorem, DF || BC. But we also have DE || BC. This cannot happen because through a given point (in this case, D), exactly one parallel can be drawn to a given line (in this case, BC).
Thus, E must be the midpoint of AC. This completes our proof.
Joining the midpoints the sides of a triangle
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:
Four smaller congruent triangles
We have:
Δ
A
D
E
≡
Δ
F
E
D
≡
Δ
B
D
F
≡
Δ
E
F
C
Proof: Consider the quadrilateral DEFB. By the midpoint theorem, we have:
1. DE = ½ BC = BF
2. DE || BF
Thus, DEFB is a parallelogram, which means that
Δ
F
E
D
≡
Δ
B
D
F
.Similarly, we can show that AEFD and DECF are parallelograms, and hence all the 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 correct).
Solved examples on the 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.
Mid point theorem example
Show that XY will bisect AD.
Solution. By the midpoint theorem, XY || BC. Now, consider
Δ
A
B
D
. 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.
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 the problem better. Consider three lines and two transversals, as shown below:
Mid point theorem example
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:
Intercepts on transversal are equal
Consider
Δ
A
C
F
