State and prove Basic proportionality theorem or Thales theorem:
A straight line drawnto one side of a triangle the other two sides, in the same ratio.
Proof:
Draw a line \(DE \parallel BC\).
Here, [Corresponding angles are equal because \(DE \parallel BC\)]
Since both triangles have a common angle,
Therefore, \(\triangle ABC \sim \triangle ADE\)
Since, corresponding sides are proportional
\(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\)
[Taking reciprocal]
Hence proved.
Answer variants:
divides the sides
\(\angle ABC = \angle ADE = \angle 1\)
\(\angle ACB = \angle AED = \angle 2\)
\(\frac{AB}{AD} = \frac{AC}{AE}\)
\(\frac{DB}{AD} = \frac{EC}{AE}\)
\(\frac{AD + DB}{AD} = \frac{AE + EC}{AE}\)
intersecting
\(\angle BAC = \angle DAE = \angle 3\)
parallel
\(\frac{AD}{DB} = \frac{AE}{EC}\)