If $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$ $a/b = b/c = c/d$ ,then how do I prove that ${(a - d)}^{2} = {(b - c)}^{2} + {(c - a)}^{2} + {(b - d)}^{2} ?$ $(a-d)^2 = (b-c)^2 +(c-a)^2 + (b-d)^2?$

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If $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$ $a/b = b/c = c/d$ ,then how do I prove that ${(a - d)}^{2} = {(b - c)}^{2} + {(c - a)}^{2} + {(b - d)}^{2} ?$ $(a-d)^2 = (b-c)^2 +(c-a)^2 + (b-d)^2?$

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Answer 1 · 2017-10-08T22:50:44

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Explanation:

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:

$b^{2} - 2 b c + c^{2} + c^{2} - 2 a c + a^{2} + b^{2} - 2 b d + d^{2}$ $b^2-2bc+c^2+c^2-2ac+a^2+b^2-2bd+d^2$

Rearranging and combining terms:

$a^{2} + 2 b^{2} + 2 c^{2} - 2 a c - 2 b c - 2 b d + d^{2}$ $a^2+2b^2+2c^2-2ac-2bc-2bd+d^2$

Factoring out the 2:

$a^{2} + 2 (b^{2} + c^{2} - a c - b c - b d) + d^{2}$ $a^2+2(b^2+c^2-ac-bc-bd)+d^2$

Now let’s use the fact that $\frac{a}{b} = \frac{b}{c}$ $a/b=b/c$

Multiply everything by $b c$ $bc$ , so:

$a c = b^{2}$ $color(red)(ac=b^2)$

Same thing for $\frac{b}{c} = \frac{c}{d}$ $b/c=c/d$ but multiply by $c d$ $cd$ , so:

$b d = c^{2}$ $color(blue)(bd=c^2)$

Same thing for $\frac{a}{b} = \frac{c}{d}$ $a/b=c/d$ but multiply by $b d$ $bd$ , so:

$a d = b c$ $color(green)(ad=bc)$

Find and replace these terms in our expression:

$a^{2} + 2 (b^{2} + c^{2} - a c - b c - b d) + d^{2}$ $a^2+2(color(red)(b^2)+color(blue)(c^2)-ac-color(green)(bc)-bd)+d^2$

$a^{2} + 2 (a c + b d - a c - a d - b d) + d^{2}$ $a^2+2(color(red)(ac)+color(blue)(bd)-ac-color(green)(ad)-bd)+d^2$

Combine like terms inside the parentheses:

$a^{2} + 2 (- a d) + d^{2} = a^{2} - 2 a d + d^{2}$ $a^2+2(-ad)+d^2=a^2-2ad+d^2$

Factor:

${(a - d)}^{2}$ $(a-d)^2$

Therefore:

${(a - d)}^{2} = {(b - c)}^{2} + {(c - a)}^{2} + {(b - d)}^{2}$ $(a-d)^2=(b-c)^2+(c-a)^2+(b-d)^2$

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If $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$ $a/b = b/c = c/d$ ,then how do I prove that ${(a - d)}^{2} = {(b - c)}^{2} + {(c - a)}^{2} + {(b - d)}^{2} ?$ $(a-d)^2 = (b-c)^2 +(c-a)^2 + (b-d)^2?$

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If ab=bc=cda/b = b/c = c/d,then how do I prove 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?

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If $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$ $a/b = b/c = c/d$ ,then how do I prove that ${(a - d)}^{2} = {(b - c)}^{2} + {(c - a)}^{2} + {(b - d)}^{2} ?$ $(a-d)^2 = (b-c)^2 +(c-a)^2 + (b-d)^2?$