Midpoint/Parallel Theorems |

**Two Theorems on Triangles and the Midpoints of Sides**

There are two theorems in Euclidean geometry that describe the relationships between the lines drawn from the midpoints of the sides of a triangle and the other sides of the triangle.

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**Line Joining Two Midpoints Theorem**

**The line which joins the midpoints of two sides of a triangle is parallel
to the third side and equal to one-half of it.**

**Given:** ABC is a triangle in which D and E are the midpoints of

sides AB and AC respectively.

**Req'd to Prove:** DE//BC** **and DE = ½ BC.

**Proof:** Produce DE to F making EF = DE. Join FC.

- In triangles ADE and CEF:

AE = EC and DE = EF (given, proof),

Angle AED = ANGLE CEF

- (SAS)

- so AD = FC (def. of congruence)

since Angle ADE = Angle EF

AB//FC

So DBCF is a parallelogram (pair of opposite sides are equal and parallel)

therefore DE//BC and DF = BC (parallelogram)

but, DE = ½ DF (construction)

therefore DE = ½ BC

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**Line from Midpoint Parallel to Another Side Theorem**

**The line drawn from the midpoint of one side of a triangle
parallel to another side, bisects the third side.**

**Given:** D is the midpoint of side AB of triangle ABC.

DE is drawn parallel to BC and meets AC at E.

**Req'd to Prove:** E is the midpoint of AC (ie: AE = EC).

**Proof:** Draw EF // AB meeting BC at F.

- Since DE // BC and EF // AB, DEFB is a parallelogram. (definition)

So EF = DB = AD (parallelogram thm.)

Now, in triangles ADE and EFC:

Angle DAE =Angle FEC and Angle AED =Angle ECF (corresponding angles)

AD = EF (proof)

- (Angle, Side, Angle)

so AE = EC and E is the midpoint of AC. (congruence)

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**Example 1**

Prove that the lines which join the midpoints of the three sides of a triangle

divide the triangle into four congruent triangles.

**Given:** In triangle ABC: D, E, and F are the midpoints of AB, AC and BC respectively.

**Req'd to Prove:** That DE, EF and DF divide triangle ABC into four congruent triangles.

**Proof:**Since D and E are mid points,

- DE = ½ BC = BF = FC (midpoint thm., definition)

Similarly, DF = AE = EC and EF = AD = DB

- So, (Side, Side, Side).

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**Example 2**

Prove that any line drawn from the vertex of a triangle to the opposite side

is bisected by the line joining the midpoints of the other two sides.

**Given:** In triangle ABC, D and E are the midpoints of AB and AC respectively.

**Req'd to Prove:** DE bisects AF, any line from A to BC.

**Proof:** Draw AF, any line from A to BC meeting BC at F and DE at G.

- DE//BC (midpoint thm.)

In triangle ABF, D is the midpoint of AB (given)

DG//BF (proof)

therefore G is the mid point of AF (midpoint thm.)

So DE bisects AF. (definition of a bisector)

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This example illustrates a **corollary** (a theorem or proposition which follows readily from another theorem or proposition) of the midpoint theorems, which states that **the line from the midpoint of one side of a triangle parallel to the base, bisects the altitude of the triangle.**

Since the line joining the midpoints of two sides of a triangle is parallel to the third side, we can state that **the line joining the midpoints of two sides of a triangle bisects the altitude of the triangle drawn to the third side.**

**Practice**

Do the following questions in a **Given, Req'd to Prove, Proof** format.

Be sure to draw a diagram and state the authority for the statements in the Proof.

1) Prove that the middle point of the hypotenuse of a right-angled triangle

is equidistant from the three vertices.

2) Through a point **D** in side **AB** of **triangle ABC**, a line is drawn parallel to **BC** meeting

**AC** in **E**. **AC** is produced to **F** so that **CF = CE**. Prove that **DF** is bisected by **BC** at G.

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**Solutions**

1) Prove that the middle point of the hypotenuse of a right-angled triangle

is equidistant from the three vertices.

**Given:** In Triangle ABC, Angle B = 90^{°}** **and D is the midpoint of AC.

**Req'd to Prove:** AD = BD = CD.

**Proof:** Draw DE // BC meeting AB at E.

- Angle AED = 90° (corresponding Angles)

AE = EB (midpoint theorem)

In triangles AED and BED:

AE = EB, Angle AED = 90° = Angle BED (proof)

DE = DE (common side)

So Triangle AED Triangle BED (Side, Angle, Side)

Therefore AD = BD (congruence)

but, AD = DC (given)

Therefore AD = BD = DC (axiom)

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2) Through a point **D** in the side **AB** of **Triangle ****ABC**, a line is drawn parallel to **BC** and meets **AC** in **E**. **AC** is produced to **F**, making **CF = CE**. Prove that **DF** is bisected by **BC**.

**Given:** D is any point in side AB of Triangle ABC. DE //BC and

intersects AC at E. AC is produced to F such that CF = CE.

**Req'd to Prove:** DF is bisected by BC. (DG = GF)

**Proof:** Join DF to intersect BC at G.

- In Triangle DEF: C is the midpoint of EF (given)

CG // DE (given)

therefore G is the midpoint of DF (midpoint thm.)

therefore BC bisects DF

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