The diagram below models the layout at a carnival where G, R, P, C, B, and E are various locations on the grounds. GRPC is a parallelogram. Parallelogram GRPC with point B between C and P forming triangle GCB where GC equals 350 ft, CB equals 300 ft, and GB equals 400 ft, point E is outside parallelogram and segments BE and PE form triangle BPE where BP equals 200 ft. Part A: Identify a pair of similar triangles. (2 points) Part B: Explain how you know the triangles from Part A are similar. (4 points) Part C: Find the distance from B to E and from P to E. Show your work. (4 points)

 A) ΔGCB ≅ ΔERG

              B) Angle-Angle (AA) Similarity Theorem

               C)  BE = 266 2/3         PE = 233 1/3

Step-by-step explanation:

GC || ER  therefore, by Alternate Interior Angles Theorem: ∠G ≡ ∠E

GR || CE  therefore, by Alternate Interior Angles Theorem: ∠C ≡ ∠R

Since we have two congruent angles, we can prove ΔGCB and ΔERG are congruent using the Angle-Angle (AA) Similarity Theorem.

Since it is a parallelogram, the sides are congruent (GC = RP and GR = CP)

GC = 350                  GB = 400                    CB = 300

ER = 350 + PE           EG = 400 + BE           RG = 300 + 200

Since the triangles are similar, their sides are proportional.