state and prove basic propotionally theore​

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state and prove basic propotionally theore​

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Ella 1 month 2021-09-10T18:04:05+00:00 2 Answers 0 views 0

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    2021-09-10T18:05:06+00:00

    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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    2021-09-10T18:05:43+00:00

    Answer:

    Basic Proportionality Theorem With Applications:

    theorem \: 1

    A line drawn parallel to one side of a triangle divides the other two sides proportionally .

    Given:

    In ABC;

    line DE is drawn parallel to side BC which meets AB at D and E.

    To prove: AD/DB=AE/EC

    PROOF:

    STATEMENT. REASON :

    In ABC and ADE,

    1. <ABC=<ADE. [Corresponding angles]
    2. <ACB=<AED. [Corresponding angles]
    3. <ABC=<ADE. [Common]

    ABC~ADE. [AAA postulate ]

    AB/DE=AE/EC. [Corresponding sides of similar triangle are proportional]

    =>AD+DE/AD=AE+EC/AE

    =>1+DB/AD=1+EC/AE

    DB/AD=EC/AE

    AD/DE=AE/EC............hence proved

    Theorem 2

    The areas of two similar triangles are proportional to the square on their corresponding sides.

    Given:

    ABC~EF such that<BAC=<EDF, <B=<E and <C=<F.

    To prove:

     \frac{area \: of \: abc}{area \: of \: def}

    Construction:

    Draw AM bisect BC and DN bisect EF.

    Proof:

    Area of ABC=1/2 BC*AM(Area of =1/2 base*altitude

    Area of DEF=1/2EF*DN(Area of =1/2 base*altitude)

    BC/EF*AM/DN......1

    IN ABM AND DEN:

    ABM~DEN. (By AA postulate)

    AM/DN=AB/DE(Corresponding sides of similar are in proportion).......2

    In ABC~DEF

    AB/DE=BC/EF=AC/DF.......3

    AM/DN=BC/EF. [FROM 2 AND 3]

     \frac{area \: of \: triangle \: abc}{area \: of \: traingle \: def}

    = BC/EF*BC*EF=BC²/EF²......4

    Now combining 3 and 4

    Area of ABC/Area of DEF

    =AB²/DE²

    =BC²/EF²

    =AC²/DF²

    hope it’s helpfull

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