Question

prove thales theorem ....​

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 sides are divided in the same ratio

Given

A ΔABC in which DE||BC and DE intersect AB and AC at D and E respectively

To Prove

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

Construction

Join BE and CD , Draw EL⊥AB and DM⊥AC

Proof

We have

ar(ΔADE) = 1/2 × AD × EL

and ar(ΔDBE) = 1/2 × DB × EL

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

Again ar(ΔADE) = ar(ΔAED)=1/2×AE×DM

and ar(ΔECD)= 1/2×EC×DM

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

Now ΔDBE and ΔECD being on the the same base DE and between the same parallels DE and BC , we have

ar(ΔDBE) = ar(ΔECD)

We get

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