Question

Can anyone give another proof of Thales Theorem ?

Answer 1

Answer:

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Answer 2

$\huge\underbrace\mathtt\color{pink}{Hlw\:Mate}$

Basic Proportionality Theorem (BPT) or Thales theorem

$\Large\bold\blue{Statement}$

A straight line drawn parallel to a side of triangle intersecting the other two sides, divides the sides in the same ratio.

$\Large\bold{\underline{\pink{Proof}}}$

$\Large\bold{\underline{\green{Given: }}}$

In Δ ABC, D is a point on AB and E is a point on AC.

$\Large\bold\purple{To \:prove:}$

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

$\Large\bold{\underline{\orange{Construction:}}}$

Draw a line DE || BC

$\Large\bold{\underline{\blue{Steps:}}}$

∠ ABC = ∠ADE =∠1 [Corresponding angles are equal because DE || BC]

∠ACB = ∠AED = ∠2 [Corresponding Angel's are equal because DE || BC]

∠DAE = ∠BAE = ∠3 [Both triangles have a common angle]

ΔABC ~ ΔADE

$\frac{AB }{AD} = \frac{AC}{AE}$

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

$1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}$

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

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

$\Large \green{\mid{\fbox{\tt Hence\:Verified }\mid}}$

$\Large\color{aqua}Happy\:Learning$

$\Large\fbox\red{Mark\:As\:Brainliest}$