Table of Contents

Last modified on September 6th, 2022

The midsegment of a triangle is a line connecting the midpoints or center of any two (adjacent or opposite) sides of a triangle. It is parallel to the third side and is half the length of the third side.

**How Many Midsegments Does a Triangle Have**

Since a triangle has three sides, each triangle has 3 midsegments. In the given ∆ABC, DE, EF, and DF are the 3 midsegments. The 3 midsegments form a smaller triangle that is similar to the main triangle. Thus, ∆ABC ~ ∆FED

- It joins the midpoints of 2 sides of a triangle; in ∆ABC, D is the midpoint of AB, E is the midpoint of AC, & F is the midpoint of BC
- A triangle has 3 possible midsegments; DE, EF, and DF are the three midsegments
- The midsegment is always parallel to the third side of the triangle; so, DE ∥ BC, EF ∥ AB, and DF ∥ AC
- The midsegment is always 1/2 the length of the third side; so, DE =1/2 BC, EF =1/2 AB, and DF =1/2 AC

The formula to find the length of midsegment of a triangle is given below:

**To prove: DE **∥** BC; DE = ½ BC**

**Proof**: A line is drawn parallel to AB, such that when the midsegment DE is produced it meets the parallel line at F

Given: D is the midpoint of AB

E is the midpoint of AC

F is the midpoint of BC

Steps | Statement | Reason |
---|---|---|

1. | In ∆ADE and ∆CFE AE = EC ∠AED = ∠CEF ∠DAE = ∠ECF | E is the midpoint of AC Vertically opposite angle Alternate angles |

2. | ∆ADE ≅ ∆CFE | By AAS congruency of triangle |

3. | DE = FE AD = CF | Corresponding parts of Congruent triangles (CPCTC) are congruent |

4. | AD = BD BD = CF | D is the midpoint of AB |

5. | DF ∥ BC and DF = BC DE ∥ BC and DF = BC DE = ½ DF | DBCF is a paralleogram |

6. | DE =1/2 BC | DF = BC Hence Proved |

**To prove**: DE is the midsegment of ∆ABC

AD = DB; AE = EC

**Proof**:

Given: D is the midpoint of AB

E is the midpoint of AC

DE ∥ BC

DE = 1/2 BC

Steps | Statement | Reason |
---|---|---|

1. | AD = DB AB = AD + DB = DB + DB = 2DB | D is the midpoint of AB |

2. | DE ∥ BC BD ∥ CF | DBCF is a parallelogram |

3. | BD = CF DA = CF | Opposite sides of a parallelogram are equal |

4. | In ∆ADE and ∆CFE, DA = CF | ∠AED = ∠CEF (Vertically opposite angle)∠DAE = ∠ECF (Alternate angles) |

5. | ∆ADE ≅ ∆CFE | By AAS congruency of triangle |

6. | AE = EC (E is the midpoint of AC) Similarly, AD = DB (D is the midpoint of AB) DE is the midsegment of ∆ABC | Corresponding parts of Congruent triangles (CPCTC) are congruent |

**Find MN in the given triangle. Given BC = 22cm, and M, N are the midpoints of AB and AC.**

Solution:

As we know, by midpoint theorem,

MN = ½ BC, here BC = 22cm

= ½ x 22 = 11cm

**Given G and H are the midpoints and GH = 17m. Find FG.**

Solution:

As we know, by the midpoint theorem,

HI = ½ FG, here HI = 17 m

FG = 2 HI = 2 x 17 = 34 m

**Solve for x in the given triangle. Given that D and E are midpoints.**

Solution:

As we know, by midpoint theorem,

DE = ½ XZ, here XZ = 32 units

3x -2 = ½ x 32

3x = 16 + 2 x = 6

Last modified on September 6th, 2022