Pokazuju da (b ^ 2-c ^ 2) * cotA + (c ^ 2-a ^ 2) * cotB + (^ 2-b ^ 2) * cotC = 0?

Po sinusnom zakonu znamo

# A / b = sina / sinB = C / sinc = 2R #

Sada

1. dio

# (B ^ 2-c ^ 2) Cota #

# = (4R ^ ^ 2sin 2B-4R ^ 2sin ^ 2C) Cota #

# = 4R ^ 2 (1/2 (1 cos2B) -1/2 (1 cos2C) Cota #

# = 4R ^ 2xx1 / 2 (cos2C-cos2B) Cota #

# = 2R ^ 2xx2sin (B + C) sin (B-C) cosa / sina #

# = 4R ^ 2sin (pi-A) sin (B-C) cosa / sina #

# = 4R ^ 2sinAsin (B-C) cosa / sina #

# = 4R ^ 2sin (B-C) # cosa

# = 4R ^ 2 (sinBcosCcosA-cosBsinCcosA) #

slično

2. dio # = (C ^ 2-a ^ 2) cotB #

# = 4R ^ 2 (sinCcosAcosB-cosCsinAcosB) #

3. dio # = (A ^ 2-b ^ 2) cotC #

# = 4R ^ 2 (sinAcosBcosC-cosAsinBcosC) #

Dodajemo tri dijela

Cijeli izraz

# (B ^ 2-c ^ 2) + Cota (c ^ 2-a ^ 2) cotB + (a ^ 2-b ^ 2) = 0 cotC #