Question

如图,三角形ABC中,AB=AC,P为BC上任意一点。求证:AB平方-AP平方=PB*PC

求证:AB平方-AP平方=PB*PC

Answer 1

做BC垂线AM，垂足M,则BM=CM
AB²＝AM²+BM²
AP²=PM²+AM²
∴AB²-AP²=AM²+BM²-PM²-AM²
=BM²-PM²
=(BM+PM)(BM-PM)
= PB*PC

Answer 2

详细解答请看下图。

Answer 3

作AD垂直BC
又因为AB＝AC
所以BD=DC
AB2=AD2+BD2
AP2=AD2+PD2
AB2-AP2=BD2-PD2
PB*PC=(BD-PD)*(PD+CD)
=(BD-PD)*(PD+BD)
=BD2-PD2
所以AB2-PD2=PB*PC

Answer 4

做BC垂线AM，垂足M,则BM=CM
AB²＝AM²+BM²
AP²=PM²+AM²
∴AB²-AP²=AM²+BM²-PM²-AM²
=BM²-PM²
=(BM+PM)(BM-PM)
= PB*PC

Answer 5

不能做出BC⊥A,M