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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#ed_op#br#ed_cl#同理,BP為∠B的中線,CQ為∠C的中線#ed_op#br#ed_cl#三角形三中線交於一點,此點為重心.
#ed_op#br#ed_cl#沒有這麼簡單...ARF並非直線, BR=/=CR, 但AB+BR=AC+CR;#ed_op#br#ed_cl##ed_op#br#ed_cl#中間的交點為Nagel Point...#ed_op#br#ed_cl#只想到用Ceva's theorem 證....但若要再證Ceva's theorem就比較麻煩...#ed_op#br#ed_cl##ed_op#br#ed_cl#
#ed_op#br#ed_cl#設BC為a, AC為b, AB為c：#ed_op#br#ed_cl#AB+BR=AC+CR=(a+b+c)/2 =&gt; BR=(a+b-c)/2, CR=(a+c-b)/2;#ed_op#br#ed_cl#同理 AP=(a+b-c)/2=BR,  AQ=(a+c=b)/2=CR, BQ=CP=(b+c-a)/2;#ed_op#br#ed_cl#據Ceva's theorem：#ed_op#br#ed_cl#
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#ed_op#br#ed_cl##ed_op#img src="http://upload.wikimedia.org/math/7/7/a/77a059835c5d8e24cd5e99864fe0bb5c.png"#ed_cl##ed_op#br#ed_cl#
#ed_op#br#ed_cl#(AQxBRxCP)/(QBxRCxPA)=(AQxBRxCP)/(CPxAQxBR)=1#ed_op#br#ed_cl#∴AR, BP, CQ are concurrent.#ed_op#br#ed_cl##ed_op#br#ed_cl#
guevara4900 寫到:圓F切∠Ａ的二邊和BC,所以BR=CR,故AR為∠Ａ中線#ed_op#br#ed_cl#同理,BP為∠B的中線,CQ為∠C的中線#ed_op#br#ed_cl#三角形三中線交於一點,此點為重心.
#ed_op#br#ed_cl#沒有這麼簡單...ARF並非直線, BR=/=CR, 但AB+BR=AC+CR;#ed_op#br#ed_cl##ed_op#br#ed_cl#中間的交點為Nagel Point...#ed_op#br#ed_cl#只想到用Ceva's theorem 證....但若要再證Ceva's theorem就比較麻煩...#ed_op#br#ed_cl##ed_op#br#ed_cl#

### [數學]幾何題..(84)

F,J,O是三角形ABC的三個旁心