Question

E is a point on side CB produced of an isosceles \(\triangle ABC\) with AB = AC. If AD \(\perp\) BC and EF \(\perp\) AC, prove that \(\triangle ABD \sim \triangle ECF\).

Hint:

In geometry proofs involving isosceles triangles, always start by using the property that angles opposite to equal sides are equal. This often provides the key to solving the problem.

Answer 1

Step 1: Understanding the Concept:
To prove that two triangles are similar, we can use one of the similarity criteria: Angle-Angle (AA), Side-Angle-Side (SAS), or Side-Side-Side (SSS). In this problem, we will use the AA similarity criterion by showing that two pairs of corresponding angles in \(\triangle ABD\) and \(\triangle ECF\) are equal.

Step 2: Detailed Explanation:
We are given the following information: \[\begin{array}{rl} \bullet & \text{\(\triangle ABC\) is an isosceles triangle with \(AB = AC\).} \\ \bullet & \text{E is a point on the side CB produced.} \\ \bullet & \text{\(AD \perp BC\), which means \(\angle ADB = 90^\circ\).} \\ \bullet & \text{\(EF \perp AC\), which means \(\angle EFC = 90^\circ\).} \\ \end{array}\] Our goal is to prove \(\triangle ABD \sim \triangle ECF\).
Proof:
In \(\triangle ABC\), since \(AB = AC\), the angles opposite to these sides are equal.
Therefore, \(\angle ABC = \angle ACB\).
Let's consider \(\triangle ABD\) and \(\triangle ECF\).
\begin{enumerate} \item \(\angle ADB = \angle EFC = 90^\circ\) (Given that AD \(\perp\) BC and EF \(\perp\) AC).
\item \(\angle ABD = \angle ABC\). Also, since E lies on CB produced, C is between B and E. The angle \(\angle ECF\) is the same as \(\angle ACB\). So, \(\angle ECF = \angle ACB\).
As established earlier, \(\angle ABC = \angle ACB\).
Therefore, \(\angle ABD = \angle ECF\).
\end{enumerate} Now we have two pairs of equal corresponding angles in \(\triangle ABD\) and \(\triangle ECF\).
By the Angle-Angle (AA) similarity criterion, we can conclude that \(\triangle ABD \sim \triangle ECF\).

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