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what is Thales theorem..pls give provement

Answer» 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 ratio.ConstructionJoin the vertex B of {tex}\\triangle{/tex}ABC 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{tex}\\perp{/tex}AC as shown in the given figure.ProofNow the area of\xa0{tex}\\triangle{/tex}APQ =\xa0{tex}\\frac {1}{2}{/tex}\xa0{tex}\\times{/tex}\xa0AP\xa0{tex}\\times{/tex}\xa0QN (Since, area of a triangle =\xa0{tex}\\frac {1}{2}{/tex}\xa0{tex}\\times{/tex}\xa0Base\xa0{tex}\\times{/tex}\xa0Height)Similarly, area of\xa0{tex}\\triangle{/tex}PBQ =\xa0{tex}\\frac {1}{2}{/tex}\xa0{tex}\\times{/tex}\xa0PB\xa0{tex}\\times{/tex}\xa0QNarea of\xa0{tex}\\triangle{/tex}APQ =\xa0{tex}\\frac {1}{2}{/tex}\xa0{tex}\\times{/tex}\xa0AQ\xa0{tex}\\times{/tex}\xa0PMAlso, area of\xa0{tex}\\triangle{/tex}QCP =\xa0{tex}\\frac {1}{2}{/tex}\xa0{tex}\\times{/tex}\xa0QC\xa0{tex}\\times{/tex}\xa0PM ...(i)Now, if\xa0we find the ratio of the area of triangles {tex}\\triangle{/tex}APQand {tex}\\triangle{/tex}PBQ, we have{tex}\\frac{\\text { area of } \\Delta A P Q}{\\text { area of } \\Delta P B Q}{/tex}\xa0=\xa0{tex}\\frac{\\frac{1}{2} \\times A P \\times Q N}{\\frac{1}{2} \\times P B \\times Q N}=\\frac{A P}{P B}{/tex}Similarly,\xa0{tex}\\frac{\\text { area of } \\Delta A P Q}{\\text { area of } \\Delta Q C P}{/tex}\xa0=\xa0{tex}\\frac{\\frac{1}{2} \\times A Q \\times P M}{\\frac{1}{2} \\times Q C \\times P M}=\\frac{A Q}{Q C}{/tex}\xa0...(ii)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 {tex}\\triangle{/tex}PBQ and {tex}\\triangle{/tex}QCP have the same area.area of {tex}\\triangle{/tex}PBQ = area of {tex}\\triangle{/tex}QCP ...(iii)Therefore, from the equations (i), (ii) and (iii) we can say that,{tex}\\frac{{AP}}{{PB}} = \\frac{{AQ}}{{QC}}{/tex}Also, {tex}\\triangle{/tex}ABC and {tex}\\triangle{/tex}APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that {tex}\\triangle{/tex}ABC\xa0{tex} \\sim {/tex}\xa0{tex}\\triangle{/tex}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.


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