Given: SR bisects angle TSQ angle T is congruent to angle Q prove: triangle RTS is congruent to RQS

SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ

∵ SR bisects angle TSQ ⇒ given

∴ ∠TSR ≅ ∠QSR

∴ m∠TSR ≅ m∠QSR

∵ ∠T ≅ ∠Q ⇒ given

∴ m∠T ≅ m∠Q

In two triangles RTS and RQS

∵ m∠T ≅ m∠Q

∵ m∠TSR ≅ m∠QSR

∵ RS is a common side in the two triangle

- By using the 4th case above

∴ Δ RTS ≅ ΔRQS ⇒ AAS postulate

Triangle RTS is congruent to RQS by AAS postulate of congruent

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