Question

If a line is drawn parallel to one side of a triangle it intersect the other two sides in distinct points prove that the other two sides are divided in same ratio

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Theorem:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points , then the other two sides are divided in the same ratio.

[ Basic Proportionality Theorem

Or Thales Theorem ]

Given:

In ∆ABC , $AB\parallelBC$ which intersects AB and AC at D and F respectively.

RTP:

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

Construction:

Join B , E and C ,D and then draw

$DM\perp AC \:and \: EN \perp AB$.

Proof:

Area of ∆ADE = $\frac{1}{2}\times AD \times EN$

Area of ∆BDE = $\frac{1}{2}\times BD \times EN$

So,ar(∆ADE)/ar(∆BDE)

=$\frac{\frac{1}{2}\times AD \times EN}{\frac{1}{2}\times BD \times EN}$

=$\frac{AD}{BD}$----(1)

Again Area of ∆ADE = $\frac{1}{2}\times AE \times DM$

Area of ∆CDE = $\frac{1}{2}\times EC \times DM$

So,ar(∆ADE)/ar(∆CDE)

=$\frac{\frac{1}{2}\times AE \times DM}{\frac{1}{2}\times EC \times DM}$

= $\frac{AE}{EC}$------(2)

Observe that ∆BDE and ∆CDE are on the same base DE and between same parallels BC and DE.

So ar(∆BDE) = ar(∆CDE) ---(3)

From (1),(2) & (3),. we have

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

Hence , proved .

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