Question

if we draw a line which is parallel to any side of a triangle and intersect the other side at different points, then this line divided these wo lines in the same ratio. ( using of thales theorem )​

Answer 1

.

To prove that:

AE

EC

=

AD

BD

Proof:

∠AED=∠ACB Corresponding angles

∠ADE=∠ABC Corresponding angles

∠EAD is common to both the triangles

⇒ΔAED∼ΔACB by AAA similarity

⇒

AE

AC

=

AD

AB

⇒

AE

AE+EC

=

AD

AD+BD

⇒

AE

EC

=

AD

BD

Hence proved

Answer 2

$\huge\bold\red{☆Answer☆}$

$Given: -$

$DE∣∣BC$

$To \: prove \: that: -$

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

$Proof: -$

$∠AED=∠ACB Corresponding angles$

$∠ADE=∠ABC  Corresponding angles$

$∠EAD is \\ \: common \: to \: both \: the \: triangles$

$⇒ΔAED∼ΔACB by AAA similarity$

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

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