Question

state and prove basic propotionally theore​

Answer 1

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 2

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

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