a) The pair of similar triangles are ΔGBC and ΔBEP
b) The triangles ΔGBC and ΔBEP are similar triangles by the condition of Angle - Angle - Angle.
c) The distance from B to E is 266.67 ft and from P to E is 233.33 ft
How to solve the similar triangles?
Part A:
The pair of similar triangles are ΔGBC and ΔBEP
Part B:
From the figure, In triangles ΔGBC and ΔBEP
=> ∠GBC ≅ ∠PBE [ Vertically opposite angles ]
=> ∠CGB ≅ ∠BEP [ Alternative angles ]
=> ∠GCB ≅ ∠BPE [ Alternative angles ]
From the above information, the triangles ΔGBC and ΔBEP are similar triangles by the condition of Angle - Angle - Angle
Part C:
Since the triangles, ΔGBC and ΔBEP are similar the ratio of corresponding angles will be equal.
=> BC/BP = GB/BE = GC/PE
From the given figure,
=> 300/200 = 400/BE = 350/PE
Take 300/200 = 400/BE
=> 3/2 = 400/BE
=> BE = 400(2)/3
=> BE = 266.67 ft
Take 300/200 = 350/PE
=> 3/2 = 350/PE
=> PE = 350(2)/3
=> PE = 233.33 ft
Therefore,
The pair of similar triangles are ΔGBC and ΔBEP
The triangles ΔGBC and ΔBEP are similar triangles by the condition of Angle - Angle - Angle.
The distance from B to E is 266.67 ftand from P to E is 233.33 ft