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.