Question

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

Answer 1

Answer:

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.

Answer 2

Answer:

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\parallelBCAB\parallelBC which intersects AB and AC at D and F respectively.

RTP:

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

DB

AD

=

EC

AE

Construction:

Join B , E and C ,D and then draw

DM\perp AC \:and \: EN \perp ABDM⊥ACandEN⊥AB .

Proof:

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

2

1

×AD×EN

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

2

1

×BD×EN

So,ar(∆ADE)/ar(∆BDE)

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

2

1

×BD×EN

2

1

×AD×EN

=\frac{AD}{BD}

BD

AD

----(1)

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

2

1

×AE×DM

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

2

1

×EC×DM

So,ar(∆ADE)/ar(∆CDE)

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

2

1

×EC×DM

2

1

×AE×DM

= \frac{AE}{EC}

EC

AE

------(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}

DB

AD

= \frac{AE}{EC}

EC

AE

Hence , proved .

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