塞瓦定理

Ceva's theorem

If the straight lines joining the vertices A，B，C of a triangle to a point O of its plane meet the opposite sides at P，Q，R，then

BP

CQ

AR

── · ── · ──＝1

PC

QA

RB

pf:ThroughO draw a line LMN parallel to BC , meeting BC at L∞ , CA at M，AB at N . Then the ranges (BPL∞C)，(QAMC) are perspective from O. Hence

BP · L∞C

BP

QA · MC

─────＝──＝────

BC · L∞P

BC

QC · MA

Similarly(CPL∞B)，(RANB) are perspective from O . Hence

CP · L∞B

CP

RA · N∞B

─────＝──＝─────

CB · L∞P

CB

RB · N∞A

⇒

BP

QA

CM

RB

AN

──＝── · ── · ── · ──

PC

CQ

MA

AR

NB

Notice that AN：NB=AM:MC,we're done.