讨论:三角形的五心

外心基本性质(庄坤霖)

已知：

AD=DB,BE=EC

AB⊥OD,BC⊥OE

求证：

(1)OA=OB=OC

(2)若AF=FC则OF⊥AC
∵AB线不平行BC∴中垂线会交于一点
∵△AOD≅△BOD(SAS)
∴AO=BO
∵△BOE≅△COE(SAS)
∴CO=BO∴OA=OB=OC
∵△AOF≅△COF(SSS)
∴∠OFC=∠OFA
∠OFC+∠OFA=180°
∠OFC+∠OFC=180°
2∠OFC=180°
∠OFC=90°
∠OFC=∠OFA=90°∴OF⊥AC
∴OF是AC的中垂线

海伦公式(庄坤霖)

已知：△ABC之三边长为a,b,c

$s={\frac {a+b+c}{2}}$

求证：

△=${\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}$
△=${\frac {1}{2}}$*a*b*sinr
=${\frac {1}{2}}ab{\sqrt {1-cos^{2}r}}$
=${\frac {1}{2}}ab{\sqrt {1-({\frac {a^{2}+b^{2}-c^{2}}{2ab}})^{2}}}$
=${\frac {1}{2}}ab{\sqrt {{\frac {(2ab)^{2}}{(2ab)^{2}}}-{\frac {(a^{2}+b-c)^{2}}{(2ab)^{2}}}}}$
=${\frac {1}{2}}ab{\sqrt {\frac {(2ab)^{2}-(a^{2}+b^{2}-c^{2})^{2}}{(2ab)^{2}}}}$
=${\frac {1}{2}}ab{\frac {\sqrt {(2ab)-(a^{2}+b^{2}-c^{2})}}{\sqrt {(2ab)^{2}}}}$
=${\frac {1}{2}}ab{\frac {\sqrt {(2ab)^{2}-(a^{2}+b^{2}-c^{2})^{2}}}{2ab}}$
=${\frac {1}{4}}{\sqrt {(2ab)^{2}-(a^{2}+b^{2}-c^{2})^{2}}}$
=${\frac {1}{4}}{\sqrt {[2ab+a^{2}+b^{2}-c^{2}][2ab^{2}-a^{2}+b^{2}-c^{2}]}}$
=${\frac {1}{4}}{\sqrt {[(a+b)^{2}-c^{2}][c^{2}-(a^{2}+b^{2}-2ab)]}}$
=${\frac {1}{4}}{\sqrt {(a+b+c)(a+b-c)[c-(a-b)][c+(a-b)]}}$
=${\sqrt {{\frac {(a+b+c)}{2}}{\frac {(a+b-c)}{2}}{\frac {(c-a+b)}{2}}{\frac {(c+a-b)}{2}}}}$