Question

prove theorem 6.2 class 10​

Answer 1

Answer:

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Theorem 6.2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Step-by-step explanation:

hey follow me and mark my answers as brainlest

Answer 2

Correct Question:-

$\odot$ Prove theroam 6.2 :- If line a divides any two sides of a triangle in the same ratio, then the line is parellel to third side.

Given:-

$\implies$ ∆ABC and a line DE intersecting AB at D and E.

Such that $\dfrac{AD}{DB} = \dfrac{AE}{EC}$

To prove:-

$\hookrightarrow$ DE || BC

Construction:-

$\hookrightarrow$ Draw DE' parallel to BC

Proof:-

Since DE' || BC

by theoram 6.1 : if a line is drawn parallel to one side of a triangle to intersecting other two sides not distinct points the other two sided are divided in the same ratio.

$\therefore \tt \dfrac{AD}{DB} = \dfrac{AE'}{E'C}$____(1)

And given that,

$\tt \dfrac{AD}{DB} = \dfrac{AE}{EC}$ _____(2)

Now,

from (1) and (2)

$\tt \dfrac{AE'}{E'} = \dfrac{AE}{EC}$

Adding 1 on both side.

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

$\hookrightarrow$ Thus, E and E' coincide

$\hookrightarrow$ DE' || BC

$\therefore$ DE || BC.

Hence, proved.

Note:-

See the figure for identification.