推 deathcustom : 我原來寫得那麼複雜就是因為要考慮到小學生甚麼都 01/16 14:09
→ deathcustom : 沒學過(相似三角形的特性、中線的特性都沒學過) 01/16 14:10
→ deathcustom : 要不然FG線段被三等份，隨便就可以算完 01/16 14:10
→ wallowes : 那我就不懂了，我連箏型都沒印象了 01/16 14:14
→ musicbox810 : d大可以寫一下怎麼證明FG被三等份嗎?謝謝 01/16 14:15
It's prima facie that FG is equally divided into 3 segments FQ,QR, and RG, and
here is the proof:
1) draw a line FC
2) in the triangle BCF, G is the mid point of BC, and P is the mid point of FC.
3) Hence, BP and FG are both medians of the triangle BCF.
4) accordingly, FG is divided into FR and RG and the ratio therebetween is 2:1.
Similarly
FQ:QG=1:2
Therefore, FQ:QR:RG = 1:1:1#
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→ musicbox810 : 謝謝d大 01/16 14:37