Question

一道数学题~求过程and解答~

答的好会提高悬赏的~~~~~

Answer 1

大题条件：△ABC是等边三角形
变式一 PE、PF、BD之间的关系为：PE=BD+PF
证明：∵S△PAB= S△ABC+ S△PAC
∴1/2·AB·PE = 1/2·AC·BD+1/2·AC·PF
∵AB=AC
∴PE=BD+PF

变式二 PE、PF、BD之间的关系为：PE=BD+PF
证明：连接PA
∵S△PAC= S△ABC+ S△PAB
∴1/2·AC·PF = 1/2·AC·BD+1/2·AB·PE
∵AC=AB
∴PF=BD+PE

Answer 2

ABC为等边或等腰
1 ∵S△PAB= S△ABC+ S△PAC
∴1/2·AB·PE = 1/2·AC·BD+1/2·AC·PF
∵AB=AC
∴PE=BD+PF
2 连接PA
∵S△PAC= S△ABC+ S△PAB
∴1/2·AC·PF = 1/2·AC·BD+1/2·AB·PE
∵AC=AB
∴PF=BD+PE
要我的吧，谢谢您啦！

Answer 3

还有其他条件吧？