Math, asked by Anonymous, 10 months ago

State and prove converse of Basic Proportionality theorem.
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Answers

Answered by Anonymous
3

Converse of Basic Proportionality Theorem. Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. Given : A Δ ABC and a line intersecting AB in D and AC in E, such that AD / DB = AE / EC.

Attachments:
Answered by sohamvashith
4

Answer:

If a straight line is drawn parallel to one side of a triangle intersecting the other two sides then it divides the two sides in the same ratio.

Step-by-step explanation:

Let ABC be the triangle.

The line l parallel to BC intersect AB at D and AC at E.

To prove  

DB

AD

​  

=  

EC

AE

​  

 

Join BE,CD

Draw EF⊥AB, DG⊥CA

Since EF⊥AB,

EF is the height of triangles ADE and DBE

Area of △ADE=  

2

1

​  

× base × height=  

2

1

​  

AD×EF

Area of △DBE=  

2

1

​  

×DB×EF

areaofΔDBE

areaofΔADE

​  

=  

2

1

​  

×DB×EF

2

1

​  

×AD×EF

​  

=  

DB

AD

​  

          ........(1)

Similarly,

areaofΔDCE

areaofΔADE

​  

=  

2

1

​  

×EC×DG

2

1

​  

×AE×DG

​  

=  

EC

AE

​  

            ......(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 have

DB

AD

​  

=  

EC

AE

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