G@ry 寫到:guevara4900 寫到:圓F切∠A的二邊和BC,所以BR=CR,故AR為∠A中線
同理,BP為∠B的中線,CQ為∠C的中線
三角形三中線交於一點,此點為重心.
沒有這麼簡單...ARF並非直線, BR=/=CR, 但AB+BR=AC+CR;
中間的交點為Nagel Point...
只想到用Ceva's theorem 證....但若要再證Ceva's theorem就比較麻煩...
設BC為a, AC為b, AB為c:
AB+BR=AC+CR=(a+b+c)/2 => BR=(a+b-c)/2, CR=(a+c-b)/2;
同理 AP=(a+b-c)/2=BR, AQ=(a+c=b)/2=CR, BQ=CP=(b+c-a)/2;
據Ceva's theorem:
Given a triangle ABC, and points D, E, and F that lie on lines BC, CA, and AB respectively, the theorem states that lines AD, BE and CF are concurrent if and only if
(AQxBRxCP)/(QBxRCxPA)=(AQxBRxCP)/(CPxAQxBR)=1
∴AR, BP, CQ are concurrent.