## state and prove basic propotionally theore​

Question

state and prove basic propotionally theore​

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

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”.

## Basic Proportionality Theorem With Applications: 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.

PROOF:

STATEMENT. REASON :

2. <ACB=<AED. [Corresponding angles]

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

## Theorem2

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.

### Toprove: ### 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] = BC/EF*BC*EF=BC²/EF²......4

Now combining 3 and 4

Area of ABC/Area of DEF

=AB²/DE²

=BC²/EF²

=AC²/DF²