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Proved Thales theorem. |
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Answer» Let ABC be the triangle.The line l parallel to BC intersect AB at D and AC at E.To prove AD\u200b/DB =AE\u200b/ECJoin BE,CDDraw EF⊥AB, DG⊥CASince EF⊥AB,EF is the height of triangles ADE and DBEArea of △ADE=1/2 × base × height=1/2\u200bAD×EFArea of △DBE= 1/2 \u200b×DB×EFareaofΔDBEareaofΔADE\u200b=1/2\u200b×AD×EF/ 1/2\u200b×DB×EF\u200b=AD/DB\u200b ........(1)Similarly,areaofΔDCEareaofΔADE\u200b= 1/2 ×AE×DG/ 1/2 ×EC×DG \u200b=AE/EC\u200b ......(2)But ΔDBE and ΔDCE are the same base DE and between the same parallel straight line BC and DE.Area of ΔDBE= area of ΔDCE ....(3)From (1), (2) and (3), we haveAD\u200b/DB =AE\u200b/ECHence proved. Basic Proportionality Theorem states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points,then the line divides those sides of the triangle in proportion.Let\xa0ABC\xa0be the triangle.The line\xa0l\xa0parallel to\xa0BC\xa0intersect\xa0AB\xa0at\xa0D\xa0and\xa0AC\xa0at\xa0E.To prove\xa0AD\u200b/DB =AE\u200b/ECJoin\xa0BE,CDDraw\xa0EF⊥AB,\xa0DG⊥CASince\xa0EF⊥AB,EF\xa0is the height of triangles\xa0ADE\xa0and\xa0DBEArea of\xa0△ADE=1/2 ×\xa0base\xa0×\xa0height=1/2\u200bAD×EFArea of\xa0△DBE= 1/2 \u200b×DB×EFareaofΔDBEareaofΔADE\u200b=1/2\u200b×AD×EF/ 1/2\u200b×DB×EF\u200b=AD/DB\u200b ........(1)Similarly,areaofΔDCEareaofΔADE\u200b= 1/2 ×AE×DG/ 1/2 ×EC×DG \u200b=AE/EC\u200b ......(2)But\xa0ΔDBE\xa0and\xa0ΔDCE\xa0are the same base\xa0DE\xa0and between the same parallel straight line\xa0BC\xa0and\xa0DE.Area of\xa0ΔDBE=\xa0area of\xa0ΔDCE ....(3)From (1), (2) and (3), we haveAD\u200b/DB =AE\u200b/ECHence proved. |
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