Question

prove theorem 6.2 of class 10​

Answer 1

Answer:

The converse of BPT says:

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: Triangle ABC in which a line ‘l’ intersects AB in D and AC in E. Such that

 AD         AE

-----  =  -----

DB         EC

To prove: DE parallel to BC

Proof: Let line ‘l’ is not parallel to BC. Then, there must be another line through D,  which is parallel to BC. Let DF parallel to BC.

Using Basic proportionality theorem, we have

   AD        AF

 ------  =  ------

  DB         FC

            AD          AF

But     ------  =  ------       (given)

            DB          EC      

                        AF           AE

Therefore        ------- =  --------      

                        FC           EC

Adding ‘1’ to both the sides, we get

       AF                AE

     ------ + 1  =  ------  + 1  

       FC                 EC

   AF + FC         AE + EC

 ----------- =  -----------

       FC                  EC

        AC           AC  

     ------  =   -------   Or   FC = EC

        FC           EC

But this is true only if F and E coincide, that is, DF coincides with DE.

Hence

DE || BC

proved

Diagram in attachment

Answer 2

Answer:

Step-by-step explanation: Theorem 6.2: Converse of BASIC PROPORTIONALITY THEOREM

Statement : If a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side.

We are given a ΔABC with DE intersecting AB and AC at D and E respectively. We also have

$\frac{AD}{DB} = \frac{AE}{EC}$

To Prove That: DE║BC

Construct: DE'║BC

Solution:

Let us assume that DE is not parallel to BC. So, now we have DE' which is parallel to BC.

By BPT, as DE'║BC

$\frac{AD}{DC} = \frac{AE'}{E'C}$

But

$\frac{AD}{DB} = \frac{AE}{EC}$

So,

\frac{AE'}{E'C}[/tex] = \frac{AE}{EC}[/tex]

Add 1 on both sides.

After taking LCM and solving we get

$\frac{AC}{E'C} = \frac{AC}{EC}\\\\Therefore,\\E'C= EC$

E' and E coincide.

Hence, ED║BC