The diagram shows triangle ABC and parallelogram BDEF. FE bisects angle DEC and angle FEC = $x^\circ$.

1. Find angle DEC using the angle bisector FE.

Since $FE$ bisects $\angle DEC$ and $\angle FEC = x^\circ$,

$\angle DEC = 2x^\circ.$

2. Use parallel lines to relate angle DEC to angle ECF.

In parallelogram $BDEF$, $DE \parallel BF$. Since $B, F, C$ are collinear, $DE \parallel CF$.

Therefore,

$\angle ECF = \angle DEC = 2x^\circ$

because they are alternate interior angles.

1. Identify one pair of equal corresponding angles using parallel lines.

In parallelogram $BDEF$, $BD \parallel EF$.

Since $A, D, B$ are collinear, $AD \parallel EF$.

Also, $A, E, C$ are collinear, so $AE$ is the same straight line as $EC$.

Hence

$\angle EAD = \angle CEF = x^\circ$

corresponding angles, since $AD \parallel EF$.

2. Identify a second pair of equal corresponding angles using parallel lines and the result from part (a).

In parallelogram $BDEF$, $DE \parallel BF$.

Since $B, F, C$ are collinear, $DE \parallel CF$.

Using the result from part (a),

$\angle ECF = 2x^\circ.$

Hence

$\angle AED = \angle ECF = 2x^\circ$

corresponding angles, since $DE \parallel CF$.

3. Conclude similarity by the AA criterion.

Since

$\angle EAD = \angle CEF$

and

$\angle AED = \angle ECF,$

two pairs of corresponding angles are equal.

Therefore,

$\triangle AED \sim \triangle ECF \quad \text{(AA similarity).}$

1. Identify the larger triangle with the same angle pattern as the two given similar triangles.

From earlier parts, $\angle BAC = \angle EAD = x^\circ$ and $\angle ACB = \angle ECF = 2x^\circ$. Hence triangle $ABC$ has the same corresponding angles as triangles $AED$ and $ECF$.

So another similar triangle is $\triangle ABC$.