有

B=2C

即

B/2-C=0

即

sin(B/2-C)=0

即

sin(B/2)cosC-cos(B/2)sinC=0

即

sin(B/2)cosC+cos(B/2)sinC

=

2sin(B/2)cosC

即

sin((B+2C)/2)

=

2sin(B/2)cosC

即

cosC

=

sin((B+2C)/2)

/

2sin(B/2)

=

cos((A-C)/2)

/

2sin(B/2)

=

sin((A+C)/2)cos((A-C)/2)

/

sinB

=

sinA+sinC

/

2sinB

=

a+c

/

2b

即

a+c

/

2b

=

a2+b2-c2

/

2ab

即

b2=ac+c2

即

b2=c(a+c)

得證

設(shè)

ΔABD與ΔACD

AD邊上的高分別為h1與h2

有

SΔABC=SΔABD+SΔSCD

即

1/2·AB·AC·sin∠BAC

=

1/2·AD(h1+h2)

即

sin∠BAC

/

AD

=

h2+h1

/

AB·AC

即

sin∠BAC/AD

=

sin∠CAD·AC+sin∠BAD·AB

/

AB·AC

即

sin∠BAC/AD

=

sin∠CAD/AB

+

sin∠BAD/AC

得證