Quadrilaterals-Activity-Mid-point theorem

Objectives

1. Understand properties of triangles – a segment connecting mid-points of two sides of a triangle will be parallel to the third side and its length will be half of the third side
2. Demonstrate that the line joining the mid-points of any two sides of a triangle is parallel to the third side and equals to half the third side

Estimated Time

One period

Prerequisites/Instructions, prior preparations, if any

Prior knowledge of point, lines, angles, parallel lines, triangles, and quadrilaterals / parallelograms

Materials/ Resources needed

Digital - Computer, Geogebra application, projector. Geogebra files- Mid-point theorem1.ggb and Mid-point theorem2.ggb

Non digital - worksheet and pencil.

Process (How to do the activity)

Work sheet

Each group member will construct on her / his notebook, using pencil, scale, protractor, and compass a triangle, with the measures provided

1. Students should plot the mid-point of two sides and connect these with a line segment. Show Mid-point theorem1.ggb step by step.
   1. They should measure the length of this segment and length of the third side and check if there is any relationship
   2. They should measure the angles formed at the two vertices connecting the third side, with the two angles formed on the two mid-points
2. Question them if there is any relationship between the two segment lengths and the measures of the two pairs of angles.
   1. Ask them why these relationships are true across different constructions.
3. Prove the theorem, using Mid-point theorem2.ggb
   1. In △ ABC, D and E are the midpoints of sides AB and AC respectively.  D and E are joined.
   2. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1/2 BC.
   3. Construction: Extend line segment DE to F such that DE = EF. Draw segment CF.
   4. Proof: In △ ADE and △ CFE AE = EC  (given)
   5. ∠AED = ∠CEF (vertically opposite angles)
   6. DE = EF   (construction)
   7. Hence △ ADE ≅ △ CFE by SAS congruence rule.
      1. Therefore, ∠ADE = ∠CFE  (by CPCT) and ∠DAE = ∠FCE (by CPCT) AD = CF (by CPCT).
4. ∠ADE and ∠CFE are alternate interior angles,
5. (AB and CF are 2 lines intersected by transversal DF).
6. ∠DAE and ∠FCE are alternate interior angles,
7. (AB and CF are 2 lines intersected by transversal AC).
   1. Therefore, AB ∥∥ CF. So - BD ∥∥ CF.
8. BD = CF (since AD = BD and it is proved above that AD = CF).
   1. Thus, BDFC is a parallelogram.
9. By the properties of parallelogram, we have DF ∥∥ BC DF = BC DE ∥∥ BC.
   1. DE = 1/2BC  (DE = EF by construction)

Evaluation at the end of the activity

1. Would this theorem apply for right angled and obtuse-angled triangles?