AB=AC (ipoteza)
PB=QC ipoteza
mas∡(PBC)=mas∡(QCA) unghiuri exterioare unor unghiuri congruente (ABC isoscel baza AB)
⇒(LUL) ΔPBA≡ΔQCA⇒AP=AQ⇔ΔAPQ isoscel, cerinta
b)PB=CQ ipoteza
BM=CN=AB/2=AC/2 (ipoteza)
mas∡(PBC)=mas∡(QCA) cf.pct a)
⇒(LUL)ΔPBM≡ΔQCN⇒MQ=PM
M, N, mijloace AB, AC (ipoteza) ⇒MN l.m.inΔABC⇒MN||PQ, ⇒MNPQtrapez
cum PM=NQ⇒MNPQ trapez isoscel⇒PN=MQ, cerinta
c)PN=MQ(pct b)
PQ=PQ
PM=NQ (pct b)
⇒(LLL)ΔNPQ≡ΔMQP⇒m∡(NPQ)=m∡(MQP)⇒ΔRPQ isoscel⇔PR=RQ
su, mai simplu MNPQ trapez isoscel (pct b)⇒PR=RQ
cu PR=RQ⇔R∈mediatoareiBC
(R, punct egal departata de capetele unui segment)
AB=AC (ipoteza)⇔A apartine mediatoarei BC
R∈mediatoarei BC
A ∈mediatoarei BC⇒AR⊥BC
