states prove thales theorem or bpt
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Answer:Thales Theorem StatementLet us now state the Basic Proportionality Theorem which is as follows: If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratioStep-by-step explanation:Let us now try to prove the basic proportionality theorem statementConsider a triangle ΔABC, as shown in the given figure. In this triangle, we draw a line PQ parallel to the side BC of ΔABC and intersecting the sides AB and AC in P and Q, respectively.Basic Proportionality TheoremAccording to the basic proportionality theorem as stated above, we need to prove:AP/PB = AQ/QCConstructionJoin the vertex B of ΔABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM⊥AC as shown in the given figure.Basic Proportionality Theorem ProofProofNow the area of ∆APQ = 1/2 × AP × QN (Since, area of a triangle= 1/2× Base × Height)Similarly, area of ∆PBQ= 1/2 × PB × QNarea of ∆APQ = 1/2 × AQ × PMAlso,area of ∆QCP = 1/2 × QC × PM ………… (1)Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we havearea of ΔAPQarea of ΔPBQ = 12 × AP × QN12 × PB × QN = APPBSimilarly, area of ΔAPQarea of ΔQCP = 12 × AQ × PM12 × QC × PM = AQQC ………..(2)According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.Therefore, we can say that ∆PBQ and QCP have the same area.area of ∆PBQ = area of ∆QCP …………..(3)Therefore, from the equations (1), (2) and (3) we can say that,AP/PB = AQ/QCAlso, ∆ABC and ∆APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that ∆ABC ~∆APQ.The MidPoint theorem is a special case of the basic proportionality theorem.According to mid-point theorem, a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.Consider an ∆ABC.Mid-point TheoremConclusionWe arrive at the following conclusions from the above theorem:If P and Q are the mid-points of AB and AC, then PQ || BC. We can state this mathematically as follows:If P and Q are points on AB and AC such that AP = PB = 1/2 (AB) and AQ = QC = 1/2 (AC), then PQ || BC.Also, the converse of mid-point theorem is also true which states that the line drawn through the mid-point of a side of a triangle which is parallel to another side, bisects the third side of the triangle.Hence, the basic proportionality theorem is proved.Converse of Basic Proportionality TheoremAccording to this theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.Converse of Basic Proportionality TheoremProofSuppose a line DE, intersects the two sides of a triangle AB and AC at D and E, such that;AD/DB = AE/EC ……(1)Assume DE is not parallel to BC. Now, draw a line DE’ parallel to BC.Hence, by similar triangles,AD/DB = AE’/E’C ……(2)From eq. 1 and 2, we get;AE/EC = AE’/E’CAdding 1 on both the sides;AE/EC + 1 = AE’/E’C +1(AE +EC)/EC = (AE’+E’C)/E’CAC/EC = AC/E’CSo, EC = E’CThis is possible only when E and E’ coincide.But, DE’ || BCTherefore, DE ||BC.Hence, proved.Solved ExamplesIn a ∆ABC, sides AB and AC are intersected by a line at D and E respectively, which is parallel to side BC. Then prove that AD/AB = AE/AC.Basic Proportionality Theorem Example Solution: Given,DE || BCSo, AD/DB = AE/ECor we can interchange the ratios as;DB/AD = EC/AENow, add 1 on both sides;(DB/AD) + 1 = (EC/AE) + 1(DB + AD)/AD = (EC + AE)/AEAB/AD = AC/ AEIf we interchange the ratios again, we get;AD/AB = AE/ACHence, proved.2. Suppose a triangle ABC, where DE is a line drawn from the midpoint of AB and ends midpoint of AC at E. AD/DB = AE/EC and ∠ADE = ∠ACB. Then prove ABC is an isosceles triangle.Solution: Given, AD/DB = AE/ECBy the converse of basic proportionality theorem, we get;DE || BCBut it is given that,∠ADE = ∠ACBHence,∠ABC = ∠ACBThe side opposite to equal angles is also equal. AB = AC