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E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\Delta ABE \sim \Delta CFB$.

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Hint: Analyze the situation with a diagram. Use the properties that the opposite angles of a parallelogram are equal, its opposite sides are parallel. And the property that two angles made by the same line with two parallel lines are equal.

Complete step-by-step answer:



Consider the parallelogram ABCD above. AD is produced up to E and BE intersects CD at F.

If we compare two triangles, $\Delta ABE$ and $\Delta CFB$:
$ \Rightarrow \angle BAE = \angle BCF{\text{ }}\left[ {\therefore {\text{ Opposite angles of a parallelogram are equal}}} \right]$
$ \Rightarrow \angle ABE = \angle CFB{\text{ }}\left[ {{\text{Angles made by the same line segment BE with parallel lines AB and CD}}} \right]$$ \Rightarrow \angle AEB = \angle CBF{\text{ }}\left[ {{\text{Angles made by the same line segment BC with parallel lines AE and CD}}} \right]$
Since all the angles of triangles ABE and CFB are the same, we can say that the triangles are similar.

So, we have:
$ \Rightarrow \Delta ABE \sim \Delta CFB$

This is the required proof.

Note: If two triangles are similar then the ratios of their corresponding sides are the same. For example, in above triangles ($\Delta ABE$ and $\Delta CFB$), since both the triangles are similar so we have:
$ \Rightarrow \dfrac{{AB}}{{CF}} = \dfrac{{AE}}{{CB}} = \dfrac{{BE}}{{FB}}$