part A.
The pair of similar triangles are:
[tex]\Delta\text{ GBC and }\Delta\text{ PBE}[/tex]
part B:
The triangles are similar because two of their angles are equal i.e:
[tex]\begin{gathered} \angle\text{ BPE =}\angle\text{ BCG} \\ \angle\text{ GBC = }\angle\text{ PBE} \end{gathered}[/tex]
Given that the shape GCPR is a parallelogram, we can identify the angles as shown below:
We can conclude that if two angles of two triangles are equal, their third angles are also equal.
Hence, the triangles are similar by AAA (angle-angle-angle).
Part C:
When two triangles are similar, the ratio of their corresponding sides is equal.
Applying this, we can find the lengths of side BE and side PE.
[tex]\begin{gathered} \frac{GB}{BE}\text{ = }\frac{PB}{CB} \\ \frac{400}{BE}\text{ = }\frac{300}{200} \\ \text{Cross}-\text{Multiply} \\ BE\text{ }\times\text{ 300 = 400 }\times\text{ 200} \\ \text{Divide both sides by 300} \\ \frac{300\text{ }\times\text{ BE}}{300}\text{ = }\frac{400\times200}{300} \\ BE\text{ = }266.67 \end{gathered}[/tex][tex]\begin{gathered} \frac{PB}{CB}\text{ = }\frac{PE}{GC} \\ \frac{200}{300}\text{ = }\frac{PE}{350} \\ \text{Cross}-\text{Multiply} \\ 300\text{ }\times\text{ PE =200 }\times\text{ 350} \\ \text{Divide both sides by 300} \\ \frac{300\text{ }\times\text{ PE}}{300\text{ }}\text{ = }\frac{200\times350}{300} \\ PE\text{ = }233.33 \end{gathered}[/tex]
Hence, the distance from B to E is 266.67 ft and the distance from P to E is 233.33 ft.
