Question

solve it ?!?!???? ....​

Answer 1

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

$\color{purple} \bf Given\:Statement :$

[ For figure refer to attachment : ]

If a line divides any two sides of a triangle in the same ratio than the line is parallel to the third side. This means $\sf \purple{\dfrac{AD}{DB}=\dfrac{AE}{EC}}$

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:ProvE:-}}}}}$

Line segment DE is parallel to BC .

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } ConstructioN:-}}}}}$

Draw a line DE' assuming it to be || to BC.

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } ProoF:-}}}}}$

This theorem is converse of Basic proportionality theorem or Thales Theorem .

As per given in Question ,

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE}{EC}$

If DE is not || to BC , then draw a line BE ' assuming it to be parallel to BC . So by Basic proportionality theorem we can say that :

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}+1=\dfrac{AE'}{E'C}+1$

$\tt:\implies \dfrac{AE+EC}{EC}=\dfrac{AE'+E'C}{E'C}$

$\tt:\implies \dfrac{AC}{EC}=\dfrac{AC}{E'C}$

$\tt:\implies \dfrac{E'C}{EC}=\cancel{\dfrac{AC}{AC} }$

$\tt:\implies \dfrac{E'C}{EC}=1$

$\underline{\boxed{\red{\tt\longmapsto\:\:E'C=EC}}}$

$\bf\qquad\qquad On\: Comparing \:them:$

We can now say that C & C' must concide on each other that is they are same points . Since DE' was parallel to BC , So DE will also be parallel to BC .

$\boxed{\green{\bf\red{\dag}\:\:Hence\:Proved.}}$

Answer 2

Answer:

Given

If a line divides any two sides of a triangle in the same ratio than the line is parallel to the third side. This means $\sf \purple{\dfrac{AD}{DB}=\dfrac{AE}{EC}}$

$\large{\underline{\underline{\red{\tt{\pink{\leadsto } To\:ProvE:-}}}}}$

Line segment DE is parallel to BC .

$\large{\underline{\underline{\blue{\tt{\purple{\leadsto } ConstructioN:-}}}}}$

Draw a line DE' assuming it to be || to BC.

$\large{\underline{\underline{\red{\tt{\orange{\leadsto } ProoF:-}}}}}$

This theorem is converse of Basic proportionality theorem or Thales Theorem .

As per given in Question ,

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE}{EC}$

If DE is not || to BC , then draw a line BE ' assuming it to be parallel to BC . So by Basic proportionality theorem we can say that :

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}+1=\dfrac{AE'}{E'C}+1$

$\tt:\implies \dfrac{AE+EC}{EC}=\dfrac{AE'+E'C}{E'C}$

$\tt:\implies \dfrac{AC}{EC}=\dfrac{AC}{E'C}$

$\tt:\implies \dfrac{E'C}{EC}=\cancel{\dfrac{AC}{AC} }$

$\tt:\implies \dfrac{E'C}{EC}=1$

$\underline{\boxed{\red{\tt\longmapsto\:\:E'C=EC}}}$

$\bf\qquad\qquad On\: Comparing \:them:$

We can now say that C & C' must concide on each other that is they are same points . Since DE' was parallel to BC , So DE will also be parallel to BC .

$\boxed{\orange{\bf\blue{\dag}\:\:Hence\:Proved.}}$

Answer 3

Answer:

Given

If a line divides any two sides of a triangle in the same ratio than the line is parallel to the third side. This means $\sf \purple{\dfrac{AD}{DB}=\dfrac{AE}{EC}}$

$\large{\underline{\underline{\red{\tt{\pink{\leadsto } To\:ProvE:-}}}}}$

Line segment DE is parallel to BC .

$\large{\underline{\underline{\blue{\tt{\purple{\leadsto } ConstructioN:-}}}}}$

Draw a line DE' assuming it to be || to BC.

$\large{\underline{\underline{\red{\tt{\orange{\leadsto } ProoF:-}}}}}$

This theorem is converse of Basic proportionality theorem or Thales Theorem .

As per given in Question ,

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE}{EC}$

If DE is not || to BC , then draw a line BE ' assuming it to be parallel to BC . So by Basic proportionality theorem we can say that :

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}+1=\dfrac{AE'}{E'C}+1$

$\tt:\implies \dfrac{AE+EC}{EC}=\dfrac{AE'+E'C}{E'C}$

$\tt:\implies \dfrac{AC}{EC}=\dfrac{AC}{E'C}$

$\tt:\implies \dfrac{E'C}{EC}=\cancel{\dfrac{AC}{AC} }$

$\tt:\implies \dfrac{E'C}{EC}=1$

$\underline{\boxed{\red{\tt\longmapsto\:\:E'C=EC}}}$

$\bf\qquad\qquad On\: Comparing \:them:$

We can now say that C & C' must concide on each other that is they are same points . Since DE' was parallel to BC , So DE will also be parallel to BC .

$\boxed{\orange{\bf\blue{\dag}\:\:Hence\:Proved.}}$

Answer 4

Answer:

Given

If a line divides any two sides of a triangle in the same ratio than the line is parallel to the third side. This means $\sf \purple{\dfrac{AD}{DB}=\dfrac{AE}{EC}}$

$\large{\underline{\underline{\red{\tt{\pink{\leadsto } To\:ProvE:-}}}}}$

Line segment DE is parallel to BC .

$\large{\underline{\underline{\blue{\tt{\purple{\leadsto } ConstructioN:-}}}}}$

Draw a line DE' assuming it to be || to BC.

$\large{\underline{\underline{\red{\tt{\orange{\leadsto } ProoF:-}}}}}$

This theorem is converse of Basic proportionality theorem or Thales Theorem .

As per given in Question ,

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE}{EC}$

If DE is not || to BC , then draw a line BE ' assuming it to be parallel to BC . So by Basic proportionality theorem we can say that :

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}+1=\dfrac{AE'}{E'C}+1$

$\tt:\implies \dfrac{AE+EC}{EC}=\dfrac{AE'+E'C}{E'C}$

$\tt:\implies \dfrac{AC}{EC}=\dfrac{AC}{E'C}$

$\tt:\implies \dfrac{E'C}{EC}=\cancel{\dfrac{AC}{AC} }$

$\tt:\implies \dfrac{E'C}{EC}=1$

$\underline{\boxed{\red{\tt\longmapsto\:\:E'C=EC}}}$

$\bf\qquad\qquad On\: Comparing \:them:$

We can now say that C & C' must concide on each other that is they are same points . Since DE' was parallel to BC , So DE will also be parallel to BC .

$\boxed{\orange{\bf\blue{\dag}\:\:Hence\:Proved.}}$

Answer 5

Answer:

Given

If a line divides any two sides of a triangle in the same ratio than the line is parallel to the third side. This means $\sf \purple{\dfrac{AD}{DB}=\dfrac{AE}{EC}}$

$\large{\underline{\underline{\red{\tt{\pink{\leadsto } To\:ProvE:-}}}}}$

Line segment DE is parallel to BC .

$\large{\underline{\underline{\blue{\tt{\purple{\leadsto } ConstructioN:-}}}}}$

Draw a line DE' assuming it to be || to BC.

$\large{\underline{\underline{\red{\tt{\orange{\leadsto } ProoF:-}}}}}$

This theorem is converse of Basic proportionality theorem or Thales Theorem .

As per given in Question ,

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE}{EC}$

If DE is not || to BC , then draw a line BE ' assuming it to be parallel to BC . So by Basic proportionality theorem we can say that :

$\tt:\implies \dfrac{AD}{DB}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}=\dfrac{AE'}{E'C}$

$\tt:\implies \dfrac{AE}{EC}+1=\dfrac{AE'}{E'C}+1$

$\tt:\implies \dfrac{AE+EC}{EC}=\dfrac{AE'+E'C}{E'C}$

$\tt:\implies \dfrac{AC}{EC}=\dfrac{AC}{E'C}$

$\tt:\implies \dfrac{E'C}{EC}=\cancel{\dfrac{AC}{AC} }$

$\tt:\implies \dfrac{E'C}{EC}=1$

$\underline{\boxed{\red{\tt\longmapsto\:\:E'C=EC}}}$

$\bf\qquad\qquad On\: Comparing \:them:$

We can now say that C & C' must concide on each other that is they are same points . Since DE' was parallel to BC , So DE will also be parallel to BC .

$\boxed{\orange{\bf\blue{\dag}\:\:Hence\:Proved.}}$