Triangle RTS is congruent to RQS by AAS postulate of congruent
Step-by-step explanation:
Let us revise the cases of congruence
- 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
Learn more:
You can learn more about the congruent in brainly.com/question/3202836
#LearnwithBrainly
Answer:
17/2
Step-by-step explanation:\
7 3/2 is equal to 17/2 because 2 x 7 is 14 and 14+3 is 17 and you keep the dem so its 17/2 and you cant simplify so it just becomes 7 3/2. i hope this helped.
Answer:
Step-by-step explanation:
Answer:
22873
Step-by-step explanation:
you have to put the value from 6-94 in n the given equation 5n+7.
for e.g= (5×6+7)+(5×7+7)+(5×8+7)+ _____ so on
6(2k-3) is the answer to this problem.
make a distributive
take the 6 out of the equation and u have remaining 2k-3. 6(2k-3)
