Question

Explain THALES theorem mathematically. (Class 10 CBSE)

And get 15 points​

Answer 1

Theorem :-

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

Given :-

A ∆ABC in which DE || BC and DE intersects AB and AC at D and E respectively

To prove :-

$\sf \: \dfrac{AD}{DB} = \dfrac{AE}{EC}$

Construction :-

Join BE and CD and Draw EL ⟂ AB and DM ⟂ AC

Proof :-

we have

$\sf \: ar( \Delta \: ADE) = \frac{1}{2} \times AD \times EL$

and

$\sf \: ar( \Delta \: DBE) = \frac{1}{2} \times DB\times EL$

$\therefore \: \: \: \: \dfrac{ \sf \: ar( \Delta \: ADE) = \frac{1}{2} \times AD \times EL}{\sf \: ar( \Delta \: DBE) = \frac{1}{2} \times DB\times EL}$

$\sf = \dfrac{AD}{DB}$

Again

$\sf \: ar( \Delta \: ADE) = ar( \Delta \: AED) = \dfrac{1}{2} \times AE \times DM$

and

$\sf \: ar( \Delta \: ECD) = \dfrac{1}{2} \times EC \times DM$

$\dfrac{\sf \: ar( \Delta \: ADE) }{ \sf \: ar( \Delta \: ECD) = } = \sf\frac{\dfrac{1}{2} \times AE \times DM}{ \sf \: \dfrac{1}{2} \times EC \times DM}$

$\sf= \dfrac{AE}{EC}$

Now ∆DBE and ∆ECD being on the same base DE and between the same parallels DE and BC , we have

ar(∆DBE) = ar(∆ECD)

From (i) , (ii) and (iii) , we have

$\sf \dfrac{AD}{DB} = \dfrac{AE}{EC}$

Answer 2

Theorem :-

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

Given :-

A ∆ABC in which DE || BC and DE intersects AB and AC at D and E respectively

To prove :-

\sf \: \dfrac{AD}{DB} = \dfrac{AE}{EC}

DB

AD

=

EC

AE

Construction :-

Join BE and CD and Draw EL ⟂ AB and DM ⟂ AC

Proof :-

we have

\sf \: ar( \Delta \: ADE) = \frac{1}{2} \times AD \times ELar(ΔADE)=

2

1

×AD×EL

and

\sf \: ar( \Delta \: DBE) = \frac{1}{2} \times DB\times ELar(ΔDBE)=

2

1

×DB×EL

\therefore \: \: \: \: \dfrac{ \sf \: ar( \Delta \: ADE) = \frac{1}{2} \times AD \times EL}{\sf \: ar( \Delta \: DBE) = \frac{1}{2} \times DB\times EL}∴

ar(ΔDBE)=

2

1

×DB×EL

ar(ΔADE)=

2

1

×AD×EL

\sf = \dfrac{AD}{DB}=

DB

AD

Again

\sf \: ar( \Delta \: ADE) = ar( \Delta \: AED) = \dfrac{1}{2} \times AE \times DMar(ΔADE)=ar(ΔAED)=

2

1

×AE×DM

and

\sf \: ar( \Delta \: ECD) = \dfrac{1}{2} \times EC \times DMar(ΔECD)=

2

1

×EC×DM

\dfrac{\sf \: ar( \Delta \: ADE) }{ \sf \: ar( \Delta \: ECD) = } = \sf\frac{\dfrac{1}{2} \times AE \times DM}{ \sf \: \dfrac{1}{2} \times EC \times DM}

ar(ΔECD)=

ar(ΔADE)

=

2

1

×EC×DM

2

1

×AE×DM

\sf= \dfrac{AE}{EC}=

EC

AE

Now ∆DBE and ∆ECD being on the same base DE and between the same parallels DE and BC , we have

ar(∆DBE) = ar(∆ECD)

From (i) , (ii) and (iii) , we have

\sf \dfrac{AD}{DB} = \dfrac{AE}{EC}

DB

AD

=

EC

AE