In the given figure, EA/EC = EB/ED, prove that △EAB ∼ △ECD.

Question

In the given figure, \(\frac{EA}{EC} = \frac{EB}{ED}\), prove that \(\triangle EAB \sim\triangle ECD\).

Answer 1

As we know that,

Converse of basic proportionality theorem states that if any line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given that,

\(\frac{EA}{EC} = \frac {EB}{ED}\)

So, \(\frac{EA}{EB} = \frac {EC}{ED}\)

In \(\triangle EAB\) and \(\triangle ECD\),

\(\frac{EA}{EB} = \frac {EC}{ED}\)

\(\angle AEB= \angle DEC\)   [Vertically opposite angles]

\(\therefore \triangle EAB \sim\triangle ECD\)  [By SAS similarity criteria]

Hence Proved.