MidSegments in Triangles - MathBitsNotebook (Geo) (2024)
Mid-Segments in Triangles MathBitsNotebook.com Topical Outline | Geometry Outline | MathBits' Teacher ResourcesTerms of Use Contact Person:Donna Roberts
The mid-segment of a triangle (also called a midline) is a segment joining the midpoints of two sides of a triangle.
"Mid-Segment Theorem": The mid-segment of a triangle, which joins the midpoints of two sides of a triangle, is parallel to the third side of the triangle and half the length of that third side of the triangle.
Examples:
1.
Given M, N midpoints. MN = 12 Find DF.
ANSWER:
2.
Given D, E midpoints. DE = 3x - 5 AB = 26 Find x.
ANSWER:
3.
Given right ΔRST. G, N, J midpoints. ST = 6; RS = 8 Find perimeter of ΔGNJ.
ANSWER:
Proof of Mid-Segment Theorem - Using Coordinate Geometry
For this proof, the diagram has been positioned in the first quadrant with one side on the x-axis to keep the algebraic computations as simple as possible, without losing the general positioning of the triangle. Be aware that other positionings are also possible.
Coordinate Geometry formulas needed for this proof:
Midpoint Formula:
Distance Formula:
Proof:
Proof of Mid-Segment Theorem - Using Similar Triangles
For this proof, we will prove ΔMFN is similar ΔDFE, by SAS for similar triangles, to obtain corresponding angles for parallel lines and establish a pair of proportional sides.
Statements
Reasons
1.
1. Given
2.
2. A mid-segment joins the midpoints of two sides of a triangle.
3.
3. Midpoint of a segment divides a segment into 2 congruent segments.
4. DM = MF; FN = NE
4. Congruent segments are segments of = length.
5.DM + MF = DF; FN + NE = FE
5. Segment Addition Postulate (or Whole Quantity)
6. MF + MF = DF; FN + FN = FE
6. Substitution
7.2MF = DF; 2FN = FE
7. Addition (or Combine Like Terms)
8.;
8. Multiplication (or Division) property of equality. [This step establishes the ratio of similitude between the two triangles.]
9.
9. Reflexive Property (or Identity Property)
10.
10. SAS for Similar Triangles: If an ∠ of one Δ is congruent to the corresponding ∠ of another Δ and the lengths of the sides including these ∠s are in proportion, the Δs are similar.
11.
11. Corresponding angles in similar triangles are congruent.
12.
12. If 2 lines are cut by a transversal such that the corresponding angles are congruent, the lines are parallel.
13.
13. Corresponding sides of similar triangles are in proportion. QED.
Proof of Mid-Segment Theorem - Using Parallelogram
For this proof, we will utilize an auxiliary line, congruent triangles and the properties of a parallelogram.
Statements
Reasons
1.
1. Given
2.
2. A mid-segment joins the midpoints of two sides of a triangle.
3. Through E draw line parallel to . Extend to intersect at M1.
3. Through a point not on a line, only one line can be drawn parallel to the given line. Parallel Postulate.
4.
4. Midpoint of a segment divides a segment into 2 congruent segments.
5.∠DFE ∠FEM1
5. If 2 parallel lines are cut by a transversal, the alternate interior angles are congruent.
6. ∠FNM ∠M1NE
6. Vertical angles are congruent.
7. ΔFNM ΔM1NE
7. ASA - If 2∠s and the included side of one Δ are congruent to the corresponding parts of another Δ, the Δs are congruent.
8.
8. CPCTC - corresponding parts of congruent triangles are congruent.
9.
9. Substitution (or Transitive property)
10. DMM1E is a parallelogram
10. A quadrilateral with one pair of sides both || and congruent is a parallelogram.
11.
11. A parallelogram is a quad. with 2 pair of opposite sides parallel.
12.
12. Opposite sides of a parallelogram are congruent.
NOTE: There-posting of materials(in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use".
"Mid-Segment Theorem": The mid-segment of a triangle, which joins the midpoints of two sides of a triangle, is parallel to the third side of the triangle and half the length of that third side of the triangle.
In triangles, medians are line segments connecting the midpoint of a side with the opposite corner.Mid-segments are line segments connecting the midpoints of two sides.
Given a triangle ABC, let's draw a line segment connecting the midpoints of two of the sides, say AB and BC. To prove the Triangle Midsegment Theorem, we need to show two things: DE = (1/2)AC. DE is parallel to AC.
If a line segment adjoins the mid-point of any two sides of a triangle, then the line segment is said to be parallel to the remaining third side and its measure will be half of the third side. DE = (1/2 * BC).
In any triangle, the lines connecting the midpoints are parallel to the opposite sides and the four triangles created by these lines and the segments of the original triangle formed by the midpoints are all congruent..
Because the midsegments are half the length of the sides they are parallel to, they are congruent to half of each of those sides (as marked). Also, this means that all four of the triangles in △ A B C , created by the midsegments are congruent by SSS.
As we know that the centroid G divides the median into 2:1 ratio. Thus, we can calculate the coordinates of point G using the section formula. Hence, the coordinates of the centroid G is G ( x , y ) = ( x 1 + x 2 + x 3 3 , y 1 + y 2 + y 3 3 ) .
Definition. Two points A(x1,y1) A ( x 1 , y 1 ) and B(x2,y2) B ( x 2 , y 2 ) can be joined to form a line segment. Taking this length as the hypotenuse of a right angled triangle ABC, the length of this line segment is found using Pythagoras' Theorem. AB2=AC2+BC2.
Address: Suite 492 62479 Champlin Loop, South Catrice, MS 57271
Phone: +9663362133320
Job: District Sales Analyst
Hobby: Digital arts, Dance, Ghost hunting, Worldbuilding, Kayaking, Table tennis, 3D printing
Introduction: My name is Kieth Sipes, I am a zany, rich, courageous, powerful, faithful, jolly, excited person who loves writing and wants to share my knowledge and understanding with you.
We notice you're using an ad blocker
Without advertising income, we can't keep making this site awesome for you.
. Extend 
to intersect at M1.
∠FEM1
