state and prove basic propotionally theore Question state and prove basic propotionally theore in progress 0 Math Ella 1 month 2021-09-10T18:04:05+00:00 2021-09-10T18:04:05+00:00 2 Answers 0 views 0

## Answers ( )

Answer:Basic Proportionality Theorem states that “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio”.

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Answer:## Basic Proportionality Theorem With Applications:

Alinedrawnparalleltoonesideofatriangledividestheothertwosidesproportionally.Given:In∆ABC;lineDEisdrawnparalleltosideBCwhichmeetsABatDandE.Toprove:AD/DB=AE/ECPROOF:STATEMENT.REASON:In∆ABCand∆ADE,<ABC=<ADE.[Correspondingangles]<ACB=<AED.[Corresponding angles]<ABC=<ADE.[Common]∆ABC~∆ADE.[AAApostulate]AB/DE=AE/EC.[Correspondingsidesofsimilartriangleareproportional]=>AD+DE/AD=AE+EC/AE=>1+DB/AD=1+EC/AEDB/AD=EC/AEAD/DE=AE/EC............henceprovedTheorem2Theareasoftwosimilartrianglesareproportionaltothesquareontheircorrespondingsides.Given:∆ABC~∆EFsuchthat<BAC=<EDF,<B=<Eand<C=<F.Toprove:Construction:DrawAMbisectBCandDNbisectEF.Proof:Areaof∆ABC=1/2BC*AM(Areaof∆=1/2base*altitudeAreaof∆DEF=1/2EF*DN(Areaof∆=1/2base*altitude)BC/EF*AM/DN......1IN∆ABMANDDEN:∆ABM~∆DEN.(ByAApostulate)AM/DN=AB/DE(Correspondingsidesofsimilar∆areinproportion).......2In∆ABC~∆DEFAB/DE=BC/EF=AC/DF.......3AM/DN=BC/EF.[FROM2AND3]=BC/EF*BC*EF=BC²/EF²......4Nowcombining3and4Areaof∆ABC/Areaof∆DEF=AB²/DE²=BC²/EF²=AC²/DF²hopeit’shelpfull