Question

1. which triangles are similar?

Explanation:

Step1: Analyze triangle \( \triangle ABF \) and \( \triangle BFC \)

We know that \( \triangle ABF \) has a right - angle at \( A \), and \( \triangle BFC \) is not a right - triangle (no right - angle mark that would make it similar to \( \triangle ABF \) in terms of AA (Angle - Angle) similarity). So \( \triangle ABF
ot\sim\triangle BFC \).

Step2: Analyze triangle \( \triangle FBC \) and \( \triangle CDB \)

For \( \triangle FBC \) and \( \triangle CDB \), \( \angle BDC=\angle FCB = 90^{\circ} \). But we cannot find another pair of equal angles to satisfy the AA similarity criterion. So \( \triangle FBC
ot\sim\triangle CDB \).

Step3: Analyze triangle \( \triangle ABF \) and \( \triangle DBC \)

Since \( AB = BE \) (marked as equal), \( BF = BF \) (common side), \( \angle BAF=\angle BDE = 90^{\circ} \), but we cannot establish the AA similarity.

Step4: Analyze triangle \( \triangle BCF \) and \( \triangle DBC \)

In \( \triangle BCF \) and \( \triangle DBC \), \( \angle BDC=\angle FCB = 90^{\circ} \), and \( \angle FBC=\angle DBC \) (common angle). By the AA (Angle - Angle) similarity criterion (\( \angle 1=\angle 2 \), \( \angle 3=\angle 4 \)), if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

Answer:

\(\triangle BCF\sim\triangle DBC\) (True), \(\triangle ABF\sim\triangle BFC\) (False), \(\triangle FBC\sim\triangle CDB\) (False), \(\triangle ABF\sim\triangle DBC\) (False)