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Proof of BPT theorem Chapter 6 similiar triangle |
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Answer» Thanks BASIC PROPORTIONALITY THEOREM PROOFIf a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.\xa0Given :\xa0In a triangle ABC, a straight line\xa0l\xa0parallel to BC, intersects AB at D and AC at E.\xa0\xa0To prove :\xa0AD/DB = AE/ECConstruction :Join BE, CD.\xa0Draw EF\xa0⊥ AB and DG\xa0⊥ CAProof :Step 1 :Because\xa0EF\xa0⊥ AB, EF is the height of the triangles ADE and DBE.\xa0Area (ΔADE) = 1/2\xa0⋅ base\xa0⋅ height = 1/2\xa0⋅ AD\xa0⋅ EFArea (ΔDBE) = 1/2\xa0⋅ base\xa0⋅ height = 1/2 ⋅ DB ⋅ EFTherefore,\xa0Area (ΔADE) /\xa0Area (ΔDBE) := (1/2\xa0⋅ AD\xa0⋅ EF) / (1/2\xa0⋅ DB\xa0⋅ EF)Area (ΔADE) /\xa0Area (ΔDBE) = AD / DB -----(1)Step 2 :\xa0Similarly, we getArea (ΔADE) /\xa0Area (ΔDCE) := (1/2\xa0⋅ AE\xa0⋅ DG) / (1/2\xa0⋅ EC\xa0⋅ DG)Area (ΔADE) /\xa0Area (ΔDCE) = AE / EC -----(2)Step 3 :\xa0But\xa0ΔDBE and\xa0ΔDCE are on the same base DE and between the same parallel straight lines BC and DE.\xa0Therefore,\xa0Area (ΔDBE) = Area (ΔDCE) -----(3)Step 4 :\xa0From (1), (2) and (3), we can obtainAD / DB = AE / ECHence, the theorem is proved.\xa0 |
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