If #a/b = b/c = c/d#,then how do I prove that #(a-d)^2 = (b-c)^2 +(c-a)^2 + (b-d)^2?#

Question

1138 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-08T22:50:44

See explanation

We want to show that:

#(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^2#

Rearranging and combining terms:

#a^2+2b^2+2c^2-2ac-2bc-2bd+d^2#

Factoring out the 2:

#a^2+2(b^2+c^2-ac-bc-bd)+d^2#

Now let’s use the fact that #a/b=b/c#

Multiply everything by #bc#, so:

#color(red)(ac=b^2)#

Same thing for #b/c=c/d# but multiply by #cd#, so:

#color(blue)(bd=c^2)#

Same thing for #a/b=c/d# but multiply by #bd#, so:

#color(green)(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^2#

#a^2+2(color(red)(ac)+color(blue)(bd)-ac-color(green)(ad)-bd)+d^2#

Combine like terms inside the parentheses:

#a^2+2(-ad)+d^2=a^2-2ad+d^2#

Factor:

#(a-d)^2#

Therefore:

#(a-d)^2=(b-c)^2+(c-a)^2+(b-d)^2#