We want to show that:
(a-d)^2=(b-c)^2+(c-a)^2+(b-d)^2(a−d)2=(b−c)2+(c−a)2+(b−d)2
Expanding the expression on the right:
b^2-2bc+c^2+c^2-2ac+a^2+b^2-2bd+d^2b2−2bc+c2+c2−2ac+a2+b2−2bd+d2
Rearranging and combining terms:
a^2+2b^2+2c^2-2ac-2bc-2bd+d^2a2+2b2+2c2−2ac−2bc−2bd+d2
Factoring out the 2:
a^2+2(b^2+c^2-ac-bc-bd)+d^2a2+2(b2+c2−ac−bc−bd)+d2
Now let’s use the fact that a/b=b/cab=bc
Multiply everything by bcbc, so:
color(red)(ac=b^2)ac=b2
Same thing for b/c=c/dbc=cd but multiply by cdcd, so:
color(blue)(bd=c^2)bd=c2
Same thing for a/b=c/dab=cd but multiply by bdbd, so:
color(green)(ad=bc)ad=bc
Find and replace these terms in our expression:
a^2+2(color(red)(b^2)+color(blue)(c^2)-ac-color(green)(bc)-bd)+d^2a2+2(b2+c2−ac−bc−bd)+d2
a^2+2(color(red)(ac)+color(blue)(bd)-ac-color(green)(ad)-bd)+d^2a2+2(ac+bd−ac−ad−bd)+d2
Combine like terms inside the parentheses:
a^2+2(-ad)+d^2=a^2-2ad+d^2a2+2(−ad)+d2=a2−2ad+d2
Factor:
(a-d)^2(a−d)2
Therefore:
(a-d)^2=(b-c)^2+(c-a)^2+(b-d)^2(a−d)2=(b−c)2+(c−a)2+(b−d)2
