Question

state and prove Thales theorem or Basic proportionality theorem​

Answer 1

Given

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

To Prove

AD/DB = AE/EC

Construction

join BE, CD and Draw EF⊥AB , DN⊥AC

Proof

ar(ΔADE)/ar(ΔBDE) = (1/2×AD × EF)/(1/2×BD×EF) = AD/BD        (i)

and

ar(ΔADE)/ar(ΔBDE) = (1/2×AE×DN)/(1/2×EC×DN) = AE/EC         (ii)

But ΔBDE and ΔCDE are on the same base DE ans between the same parallels DE and BC

ar(ΔBDE) = ar(ΔCDE)                       (iii)

ar(ΔADE)/ar(ΔBDE) = ar(ΔADE)/ar(ΔCDE)                       Using (iii)         (iv)

Hence

AD/DB = AE/EC

Answer 2

Given:

DE || BC

Construction :

Draw EL ⊥ AB and ,⊥ AC and Join B to E and D to C

To Prove:

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

Proof:

We know that

$Area \: of \: triangle = \frac{1}{2} \times b \times h$

$ar (ᐃADE) =\frac{1}{2} × AD × EL$

$ar (ᐃDBE) = \frac{1}{2} × DB × EL$

$=> \frac{ar (ᐃADE)}{ar (ᐃDBE)}= \frac{½ × AD × EL}{½ × DB × EL} = \frac{AB}{DE}—(i)$

Again,

$ar(∆DEC) = \frac{1}{2} \times \: EC \times \: DM$

$= > \frac{ar(∆ADE) }{ar(∆DEC)} = \frac{ \frac{1}{2} \times AC × DM \: }{ \frac{1}{2} \times EC× DM } = \frac{AC }{EC} ..(ii)$

ar (∆BDE) = ar(∆DEC). . . . . (iii) ( both Triangles are on the same base and between same parallels

From eq (ii) and (iii)

$= > \frac{ar(∆ADE )}{ar(∆BDE)} = \frac{AE}{EC}$

From eq (i) and (iv)

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

Hence proved