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| === Objectives === | | === Objectives === |
− | 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 | + | # 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 |
| + | # 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 |
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| === Estimated Time === | | === Estimated Time === |
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| === Prerequisites/Instructions, prior preparations, if any === | | === Prerequisites/Instructions, prior preparations, if any === |
| + | Prior knowledge of point, lines, angles, parallel lines, triangles, and quadrilaterals / parallelograms |
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| === Materials/ Resources needed === | | === Materials/ Resources needed === |
− | Geogebra file | + | Digital - Computer, Geogebra application, projector. Geogebra file - Mid-point theorem.ggb |
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| + | Non digital -worksheet and pencil. |
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| === Process (How to do the activity) === | | === Process (How to do the activity) === |
| + | Work shee - t |
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| + | Each group member will construct on her / his notebook, using pencil, scale, protractor, and compass a triangle, with the measures provided |
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| + | # Students should plot the mid-point of two sides and connect these with a line segment. |
| + | ## They should measure the length of this segment and length of the third side and check if there is any relationship |
| + | ## They should measure the angles formed at the two vertices connecting the third side, with the two angles formed on the two mid-points |
| + | # Question them if there is any relationship between the two segment lengths and the measures of the two pairs of angles. |
| + | ## Ask them why these relationships are true across different constructions. |
| + | # Prove the theorem |
| + | ## In △ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. |
| + | ## Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1/2 BC. |
| + | ## Construction: Extend line segment DE to F such that DE = EF. Draw segment CF. |
| + | ## Proof: In △ ADE and △ CFE AE = EC (given) |
| + | ## ∠AED = ∠CEF (vertically opposite angles) |
| + | ## DE = EF (construction) |
| + | ## Hence △ ADE ≅ △ CFE by SAS congruence rule. |
| + | ### Therefore, ∠ADE = ∠CFE (by CPCT) and ∠DAE = ∠FCE (by CPCT) AD = CF (by CPCT). |
| + | # ∠ADE and ∠CFE are alternate interior angles, |
| + | # (AB and CF are 2 lines intersected by transversal DF). |
| + | # ∠DAE and ∠FCE are alternate interior angles, |
| + | # (AB and CF are 2 lines intersected by transversal AC). |
| + | ## Therefore, AB ∥∥ CF. So - BD ∥∥ CF. |
| + | # BD = CF (since AD = BD and it is proved above that AD = CF). |
| + | ## Thus, BDFC is a parallelogram. |
| + | # By the properties of parallelogram, we have DF ∥∥ BC DF = BC DE ∥∥ BC. |
| + | ## DE = 1/2BC (DE = EF by construction) |
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| === Evaluation at the end of the activity === | | === Evaluation at the end of the activity === |
− | | + | # Would this theorem apply for right angled and obtuse-angled triangles? |
| [[Category:Quadrilaterals]] | | [[Category:Quadrilaterals]] |
| [[Category:Class 9]] | | [[Category:Class 9]] |