In Δ BAT and Δ BAC, given triangles are congruent so the corresponding parts are: B *↔* B, A *↔* A, T *↔* C

Thus, Δ BCA ≅ Δ BTA [By SSS congruence rule]

InΔ QRS and Δ TPQ, given triangles are congruent so the corresponding parts are: P *↔* R, T *↔* Q, Q *↔* S

Thus, Δ QRS ≅ Δ TPQ [By SSS congruence rule]