The midpoint theorem states that if a segment is formed by connecting the midpoints of two of the sides of a triangle, then that segment must be parallel to the third side and equal to third side of the triangle
Given: A triangle ABC in which D is the midpoint of AB and E is the midpoint of AC.
- To Prove: DE ∥ BC and DE = 1/2(BC)
- Construction: Extend the line segment joining points D and E to F such that DE = EF, and join CF.
Proof:
In ∆AED and ∆CEF:
- DE = EF (construction)
- ∠1 = ∠2 (vertically opposite angles)
- AE = CE (E is the mid-point)
By SAS criterion, △AED ≅ △CEF.
Therefore, ∠3 =∠4 (by CPCT)
But these are alternate interior angles, so AB ∥ CF
- AD = CF (CPCT)
- But AD = DB (D is the mid-point), so BD = CF
In Quadilateral BCFD:
- BD∥ CF (since AB ∥ CF)
- BD = CF
BCFD is a parallelogram as one pair of opposite sides is parallel and equal.
Therefore,
DF∥ BC (opposite sides of parallelogram)
DF = BC (opposite sides of parallelogram)As DF∥ BC, DE∥ BC and DF = BC
But DE = EF
So, DF = 2(DE)
2(DE) = BC
DE = 1/2(BC)Hence, proved that the line joining the mid-points of two sides of the triangle is parallel to the third side and is half of it.
Midpoint Formula
The midpoint of any line segment is defined as the coordinate point that divides the line segment into two equal parts.
Suppose P(x1, y1) and Q(x2, y2) are the coordinates of the endpoints of any line segment; then the midpoint formula is given as
Midpoint = [\frac{(x_1 + x_2)}{2}, \frac{(y_1 + y_2)}{2}]
Converse of Midpoint Theorem Proof
The line drawn through the midpoint of one side of a triangle parallel to the base of a triangle bisects the third side of the triangle.
- Given: In △PQR, S is the midpoint of PQ, and ST ∥ QR.
- To Prove: T is the midpoint of PR.
- Construction: Draw a line through R parallel to PQ and extend ST to U.
Proof:
- ST∥ QR (given)
- So, SU∥ QR
- PQ∥ RU (construction)
Therefore, Quadilateral SURQ is a parallelogram.
- SQ = RU (Opposite sides of parallelogram)
- But SQ = PS (S is the mid-point of PQ)
- Therefore, RU = PS
In △PST and △RUT
- ∠1 =∠2 (vertically opposite angles)
- ∠3 =∠4 (alternate angles)
- PS = RU (proved above)
△PST ≅ △RUT by AAS criteria
Therefore, PT = RT
T is the mid-point of PR.