Question

PQRS is a quadrilateral and T and U are respectively points on PS and RS such that PQ = RQ, ∠PQT = ∠RQT and ∠TQS = ∠UQS. Prove that QT = QU.


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Answer 1

SOLUTION:-

Given:

PQ= RQ,

∠PRT= ∠RQU &

∠TQS= ∠UQS

To prove:

QT= QU

Proof:

In ∆PQS & ∆RQS, we have

PQ= RQ [given]

QS=QS [common]

Therefore,

∠PQT= ∠RQU & ∠TQS= ∠UQS

=) ∠PQT + ∠TQS = ∠RQU + ∠UQS

=) ∠PQS= ∠RQS

So,

∆PQS ≅∆RQS [S.A.S congurency]

=) ∠PSQ =∠QSR [c.p.c.t]

or ∠TSQ = ∠USQ

Again,

In ∆TQS & ∆SQU, we have

∠TQS = ∠SQU [given]

QS = QS [common]

& ∠TSQ = ∠QSU

=) ∆TQS ≅∆SQU [A.S.A congurency]

=) QT= QU [c.p.c.t]

Hence, Proved

Hope it helps ☺️

Answer 2

Answer:

SOLUTION:-

Given:

PQ= RQ,

∠PRT= ∠RQU &

∠TQS= ∠UQS

To prove:

QT= QU

Proof:

In ∆PQS & ∆RQS, we have

PQ= RQ [given]

QS=QS [common]

Therefore,

∠PQT= ∠RQU & ∠TQS= ∠UQS

=) ∠PQT + ∠TQS = ∠RQU + ∠UQS

=) ∠PQS= ∠RQS

So,

∆PQS ≅∆RQS [S.A.S congurency]

=) ∠PSQ =∠QSR [c.p.c.t]

or ∠TSQ = ∠USQ

Again,

In ∆TQS & ∆SQU, we have

∠TQS = ∠SQU [given]

QS = QS [common]

& ∠TSQ = ∠QSU

=) ∆TQS ≅∆SQU [A.S.A congurency]

=) QT= QU [c.p.c.t]

Hence, Proved

Hope it helps ☺️❤✌