Question

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

Answer 1

Explanation:

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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Answer 2

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