Maria correctly concluded that not enough information was provided to construct the requested proof. What extra information would allow Maria to contrast the proof? SA=TR. RS= ST. ASCT is similar to ASCA. SC=5/2BR

Hello your question is incomplete attached below is the complete question

RS = ST ( option b )

Step-by-step explanation:

First we consider Δ SCA and Δ TCB

∠ SCA  = ∠ TCB  ,  ∠CAS = ∠CBT = 90°

if we assume  RS = ST then ΔSRT will be termed an Isosceles triangle with its legs : SR = ST. also vertex S bisects the base of the triangle.

Δ SBR ≅ ΔSBT  from here we can deduce that ; BR = BT

From the assumptions what is actually missing to contrast the proof is : RS = ST

i dont know i just need points