Question

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Answer 1

Step-by-step explanation:

∠PQS = ∠PQT + ∠TQS

∠RQS = ∠RQU + ∠UQS

Given that ∠PQT = ∠RQU & ∠TQS = ∠UQS

=> ∠ PQS = ∠RQS

now in ΔRQS & Δ PQS

PQ = RQ given

∠ PQS = ∠RQS

QS is common

=> ΔRQS ≅ Δ PQS

=> ∠QRU = ∠QPT

now in Δ RQU  & Δ PQT

RQ = PQ

∠RQU = ∠PQT

∠QRU = ∠QPT

=> Δ RQU  ≅ Δ PQT

=> QU = QT

Hope it helps!

Answer 2

GIVEN:-

• PQ= PR

• $\rm{\angle{PQT}=\angle{RQU}}$....1

• $\rm{\angle{TQS}=\angle{UQS}}$.....2

TO PROVE:-

• QT = QU

CONGURENCE CRITERIA USED:-

• SAS (SIDE-ANGLE-SIDE).

• ASA(ANGLE-SIDE-ANGLE)

Now,

Adding eq 1 and 2.

$\rm{\angle{PQT}+\angle{TQS}=\angle{RQU}+\angle{UQS}}$

$\rm{\angle{PQS}=\angle{RQS}}$.

Now, in ∆PQS & ∆ RQS.

$\implies\rm{PQ=RQ}$(Given)

$\implies\rm{\angle{PQS}=\angle{RQS}}$.(Proved above)

$\implies\rm{QS=SQ}$ (Common sides).

Hence, It is Congurent by S-A-S Criteria.

By CPCT:-

$\implies\sf{\angle{P}=\angle{R}}$.

Therefore, In ∆ PQT & ∆ RQU.

$\implies\rm{\angle{P}=\angle{R}}$.(By CPCT)

$\implies\rm{\angle{PQT}=\angle{RQU}}$.(GIVEN).

$\implies\rm{PQ=RQ}$(Given)

So, it is Congurent By A-S-A Congurence Criteria.

Hence, QT= QU prove by CPCT.

MORE TO KNOW:-

• SSS Criteria- When all the sides of triangles are equal to the corresponding triangle.

• RHS Criteria - When the both triangle are Right angled, Hypotenuse are equal and one of the side is equal to one another.