Question

E is a point on the extended part of the side AD of a parallelogram ABCD and BE intersects CD at F; then prove that \(\triangle ABE \sim \triangle CFB\).

Hint:

When dealing with geometry proofs involving parallelograms, always list the key properties: opposite sides are parallel, opposite sides are equal, opposite angles are equal, and diagonals bisect each other. These properties are the tools you will use to find equal angles or sides.

Answer 1

Step 1: Understanding the Concept:
To prove that two triangles are similar, we can use the Angle-Angle (AA) similarity criterion. This involves finding two pairs of corresponding angles that are equal. We will use the properties of parallelograms, such as opposite angles being equal and opposite sides being parallel.

Step 2: Detailed Explanation:
Given: ABCD is a parallelogram. E is a point on the extension of side AD. BE intersects CD at F.
To Prove: \(\triangle ABE \sim \triangle CFB\).
Proof:
In \(\triangle ABE\) and \(\triangle CFB\): \begin{enumerate} \item Angle A and Angle C: In a parallelogram, opposite angles are equal. Therefore, \(\angle DAB = \angle BCD\). This means \(\angle EAB = \angle FCB\). So we have one pair of equal angles.
\item Alternate Interior Angles: Since ABCD is a parallelogram, AD is parallel to BC. As E lies on the extension of AD, the entire line AE is parallel to BC (\(AE \parallel BC\)). Now, consider BE as a transversal line intersecting these two parallel lines. The alternate interior angles formed are equal. Therefore, \(\angle AEB = \angle CBF\). So we have a second pair of equal angles.
\end{enumerate} Since two pairs of corresponding angles of \(\triangle ABE\) and \(\triangle CFB\) are equal, the third pair must also be equal (\(\angle ABE = \angle CFB\)).
By the AA similarity criterion, \(\triangle ABE \sim \triangle CFB\).

Step 3: Final Answer:
Hence, it is proved that \(\triangle ABE\) is similar to \(\triangle CFB\).