Question

proof Basic Proportionality Theorem (BPT)​

Answer 1

Answer:

PG NUMBER 124 IN NCERT TEXTBOOK

Step-by-step explanation:

We need to prove that

AD AE

DB EC = .

Let us join BE and CD and then draw DM ⊥ AC and

EN ⊥ AB.

Now, area of Δ ADE (=

1

2 base × height) =

1

2 AD × EN.

Recall from Class IX, that area of Δ ADE is denoted as ar(ADE).

So, ar(ADE) =

1

2 AD × EN

Similarly, ar(BDE) =

1

2 DB × EN,

ar(ADE) =

1

2 AE × DM and ar(DEC) =

1

2 EC × DM.

Therefore, ar(ADE)

ar(BDE) =

1 AD × EN

2 AD

1 DB DB × EN

2

= (1)

and ar(ADE)

ar(DEC) =

1 AE × DM

2 AE

1 EC EC × DM

2

= (2)

Note that Δ BDE and DEC are on the same base DE and between the same parallels

BC and DE.

So, ar(BDE) = ar(DEC)

Answer 2

Answer:

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