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.
 * 2) They should measure the length of this segment and length of the third side and check if there is any relationship
 * 3) They should measure the angles formed at the two vertices connecting the third side, with the two angles formed on the two mid-points
 * 4) Question them if there is any relationship between the two segment lengths and the measures of the two pairs of angles.
 * 5) Ask them why these relationships are true across different constructions.
 * 6) Prove the theorem, using Mid-point theorem2.ggb
 * 7) In △ ABC, D and E are the midpoints of sides AB and AC respectively.  D and E are joined.
 * 8) Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1/2 BC.
 * 9) Construction: Extend line segment DE to F such that DE = EF. Draw segment CF.
 * 10) Proof: In △ ADE and △ CFE AE = EC  (given)
 * 11) ∠AED = ∠CEF (vertically opposite angles)
 * 12) DE = EF   (construction)
 * 13) Hence △ ADE ≅ △ CFE by SAS congruence rule.
 * 14) Therefore, ∠ADE = ∠CFE  (by CPCT) and ∠DAE = ∠FCE (by CPCT) AD = CF (by CPCT).
 * 15) ∠ADE and ∠CFE are alternate interior angles,
 * 16) (AB and CF are 2 lines intersected by transversal DF).
 * 17) ∠DAE and ∠FCE are alternate interior angles,
 * 18) (AB and CF are 2 lines intersected by transversal AC).
 * 19) Therefore, AB ∥∥ CF. So - BD ∥∥ CF.
 * 20) BD = CF (since AD = BD and it is proved above that AD = CF).
 * 21) Thus, BDFC is a parallelogram.
 * 22) By the properties of parallelogram, we have DF ∥∥ BC DF = BC DE ∥∥ BC.
 * 23) 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?