Question

In the given figure PQ = QR and x = y. Prove that AR = PB

Answer 1

here is your answer OK

PSR=x∠QSR + ∠PSR =180° Linear Pair∠QSR =180°-x∠REP=y∠QEP + ∠REP =180°∠QEP =180°-y=180°-x Since x=yIn ∆ QEP and ∆ QSR∠QEP = ∠QSR Proved above∠PQE=∠RQS CommonPQ=QR GivenBy AAS Congruence∆ QEP≅∆ QSRSo,PE=RS ( CPCT) ans

I HOPE I HELP YOU

Answer 2

Answer:

set up your proof like this:

Statements | Reasons

PQ=QR |Given

m<x=m<y | Given

m<PQR=m<RQP | Reflexive Property

ΔPQB ≅ ΔRQA | ASA

(Technically you should put a step here saying line AR is congruent to line PB by CPCTC. This would make the reasons for the next step that congruent lines have equal measures.)

AR = PB | CPCTC (Corresponding Parts of Congruent Triangles are Congruent)