1. **Problem Statement:** We have a parallelogram GRPC with points B and E such that triangle GCB and triangle BPE are formed. Given lengths: $GC=375$ ft, $CB=325$ ft, $GB=425$ ft, and $BP=225$ ft. We need to: - Part A: Identify a pair of similar triangles. - Part B: Explain why these triangles are similar. - Part C: Find distances $BE$ and $PE$. 2. **Part A: Identify Similar Triangles** The triangles to consider are $\triangle GCB$ and $\triangle BPE$. 3. **Part B: Reason for Similarity** - Since GRPC is a parallelogram, opposite sides are parallel and equal. - $GC$ is parallel to $PR$ and $CB$ is parallel to $GP$. - Angles at $C$ and $P$ are corresponding angles, so $\angle GCB = \angle BPE$. - $\angle GBC = \angle PBE$ because they are vertical angles. - By AA (Angle-Angle) similarity criterion, $\triangle GCB \sim \triangle BPE$. 4. **Part C: Find $BE$ and $PE$** Since $\triangle GCB \sim \triangle BPE$, corresponding sides are proportional: $$\frac{GC}{BP} = \frac{CB}{BE} = \frac{GB}{PE}$$ Given: $$GC=375, \quad CB=325, \quad GB=425, \quad BP=225$$ Set up ratios: $$\frac{375}{225} = \frac{325}{BE} = \frac{425}{PE}$$ Calculate the scale factor: $$\frac{375}{225} = \frac{5}{3}$$ Find $BE$: $$\frac{5}{3} = \frac{325}{BE} \implies BE = \frac{325 \times 3}{5} = 195$$ Find $PE$: $$\frac{5}{3} = \frac{425}{PE} \implies PE = \frac{425 \times 3}{5} = 255$$ **Final answers:** $$BE = 195 \text{ ft}, \quad PE = 255 \text{ ft}$$