Similarity proof in an extended parallelogram

Similarity proof in an extended parallelogram

Secondary 2
TGM Original Questions

In the figure, \(ABCD\) is a parallelogram, \(E\) lies on \(DC\), and \(AE\) and \(BC\) are produced to meet at \(F\). Given \(DE:EC=2:3\):

  1. name a triangle similar to \(ADE\) and explain
  2. show that \(BC=\frac25BF\).

Solution:

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Answer:(i) \(\triangle FCE\), AAA; (ii) \(BC/BF=2/5\).

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