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| Converse of Theorem 7 |
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| A straight line drawn through the mid-point of one side and parallel to another side of a triangle bisects the third side. |
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 ABC in which D is the mid-point of AB and DE||BC. |
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| E is the mid-point of AC. i.e., to prove AE=EC. |
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| Since DE||BC, you can complete a parallelogram with DB and BC as consecutive sides. |
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| Hence draw EF||BD to meet DE produced at F. |
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| In quadrilateral DBCF, |
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| DB||CF (by construction) |
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| DF||BC (given) |
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 DBCF is a parallelogram. |
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 DB=CF ----(i) |
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| (opposite sides of a parallelogram) |
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| But DB=AD ----(ii) (given) |
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| From (i) and (ii), AD=CF |
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| Now compare triangles, AED and CEF, |
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| AD = CF |
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| \ AE = EC (CPCT) |
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| That is E is the mid point of AC. |
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| Hence the theorem is proved. |
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| Recall the above theorem and apply it to the diagram given. |
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| In the diagram if D is the mid-point of AB and DE is drawn parallel to BC, then E Will be the midpoint of AC i.e., AE=EC. |
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| Now if AX is drawn parallel to BC, then also AE=EC if AD=DB. |
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| Now draw three parallel lines AB, CD, EF as shown in the figure. |
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| Draw a transversal t1 such that AB=BC. |
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| Now draw another transversal t2. |
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| Measure DE and EF. You will find that DE=EF and AB=BC. |
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| In the diagram, AD, BE and CF are three parallel lines. |
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| AB and BC are equal intercepts made on t1. |
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| If any transversal is drawn, the intercepts made on it will also be equal. |
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