就 一視頻 一網(wǎng)友 一評論 之 具體過程 饗以諸君
有
A/2+B+(C-B）/2=π/2
即
sin∠ABP=sin(A/2+B)=cos((C-B）/2)
即
2Rcos((B-C）/2)=AP
即
AO
=
AP-OP
=
AP-BP
=
2Rcos((B-C）/2)-2Rcos((B+C）/2)
=
4Rsin(B/2)sin(C/2)
有
PC=PO=PB
即
a/OP
=
a/PC
=
2cos∠PCB
=
2cos∠PAB
=
(b+c-a)/AO
即
AO/OP=(b+c-a)/a
即
AO/OP=(sinB+sinC-sinA)/sinA
即
OP
=
4Rsin(B/2)sin(C/2)sinA
/
(sinB+sinC-sinA)
=
2Rsin(A/2)(4sin(B/2)sin(C/2)cos(A/2)
/
(sinB+sinC-sinA)
=
2Rsin(A/2)
(2cos((B-C）/2)-2cos((B+C）/2)）
cos(A/2)
/
(sinB+sinC-sinA)
=
2Rsin(A/2)
(2cos((B-C）/2)-2cos((B+C）/2)）
sin((B+C）/2)
/
(sinB+sinC-sinA)
=
2Rsin(A/2)
(sinB+sinC-sin(B+C)）
/
(sinB+sinC-sinA)
=
2Rsin(A/2)
(sinB+sinC-sinA）
/
(sinB+sinC-sinA)
=
2Rsin(A/2)
且
cosA+cosB+cosC-1
=
ab2+ac2-a3+a2b+bc2-b3+a2c+b2c-c3-2abc
/
2abc
=
(4a2b2-(a2+b2-c2）2）/(a+b+c)
/
2abc
=
2ab/(c(a+b+c）)
(4a2b2-(a2+b2-c2）2）/(4a2b2)
=
absinC/(a+b+c)
/
c/(2sinC）
=
r/R
即
AO·OP
=
8R2sin(A/2)sin(B/2)sin(C/2)
=
2R
4Rsin(A/2)sin(B/2)sin(C/2)
=
2R
2R
(cos((A-B）/2)-cos((A+B）/2)）
sin(C/2)
=
2R
2R
(cos((A-B）/2)-cos((A+B）/2)）
cos((A+B）/2)
=
2R
R
(cosA+cosB-cos(A+B)-1）
=
2R
R
(cosA+cosB+cosC-1）
=
2Rr
