Question

State and prove the converse of BPT theorem.

⛳No spammed answers !!
⛳Figure is a must !!
⛳Clear answer should be given !!​

Answer 1

R.E.F. the above two attachments

Converse of basic proportionality Theorem

Statement :

If a line divide any two sides of a triangle (∆) in the same ratio , then the line must be parallel (||) to third side.

If AD / DE = AE / EC then DE||BC.

Prove that: DE||BC.

Given in ∆ ABC , D and E are two points of AB and AC respectively , such that ,

AD / DB = AE / EC _______ (1)

Let us assume that in ∆ ABC , the point F is an intersect on the side AC. so , we can apply the

Thales Theorem,

AD / DB = AF / FC _______(2)

Simplify (1) and (2)

AE / EC = AF / FC

Adding 1 on both sides

AE / EC + 1 = AF / FC + 1

➪ AE + EC / EC = AF + FC / FC

➪AC / EC = AF / FC

➪AC / FC

From the above we can said that the points E and F are coincide on AC , i.e.., DF coincides with DE . Since DF is parallel to BC , DE is also parallel to BC .

∴ Hence , the Converse of basic proportionality Theorem is proved .

Answer 2

$\huge{\boxed{\textsf{\textbf{\color{teal}{QuesTion:-}}}}}$

State and prove the converse of BPT theorem.

$\huge{\boxed{\textsf{\textbf{\color{teal}{SoluTion:-}}}}}$

✍️Statement :-

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 :-

∆ABC and a line DE intersecting AB at D and AC at E, such that :

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

✍️To prove:-

We have to prove that DE || BC.

✍️Construction:-

Draw DE' parallel to BC.

✍️Proof:-

Since DE' || BC,

By theorem 6.1,

By theorem 6.1, If a line is drawn parallel to one side of a triangle to intersect other two sides not distinct points, the other two sides are divided in the same ratio.

$= = > \frac{AD}{DB} = \frac{AE'}{E'C} ....( \: 1 \: )$

And given that,

$= = > \frac{AD}{DB} = \frac{AE}{EC}....( \: 2 \: )$

From ( 1 ) and ( 2 ),

$= = > \frac{AE'}{E'C} = \frac{AE}{CE}$

Adding 1 on both sides,

$\frac{AE'}{E'C} +1 = \frac{AE}{EC} + 1$

$\frac{AE'+E'C}{E'C} = \frac{AC+EC}{EC}$

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

$\frac{1}{E'C} = \frac{1}{EC}$

$EC=E'C$

Thus, E and E' coincide.

Since, DE' || BC.

Therefore, DE || BC.

Hence proved !!