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Given that sides $AC$ and $A'C'$ are parallel and lengths $AB$ and $BC$ are known. Also the distance between $AC$ and $A'C'$ is known and is $w$, What I would like to know is the lengths $AA'$ and $CC'$ and also the difference between AC and A'C'

Appreciate any pointers. Thank you for your help.

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Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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  • $\begingroup$ Also we need to find the difference AC - A'C' $\endgroup$ – Kumar Jul 12 at 13:52
  • $\begingroup$ this might help $\endgroup$ – Lozenges Jul 12 at 14:22
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Let's assume $AB=a$, $BC=b$, $A'B=a'$, $BC'=b'$ and $A'D=C'D'=w$.

Clearly triangle $ABC$ and $ADA'$ are similar. Therefore,

$$\dfrac{AA'}{AC}=\dfrac{A'D}{BC}$$

Or

$$\dfrac{AA'}{\sqrt{a^2+b^2}}=\dfrac{w}{b}$$

$$AA'=\dfrac{w}{b}\sqrt{a^2+b^2}$$

Similarly,

Triangle $BCA$ and $D'CC'$ are similar. Therefore,

$$\dfrac{CC'}{AC}=\dfrac{C'D'}{AB}$$

Or

$$\dfrac{CC'}{\sqrt{a^2+b^2}}=\dfrac{w}{a}$$

$$CC'=\dfrac{w}{a}\sqrt{a^2+b^2}$$

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Akash Karnatak is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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  • $\begingroup$ Thank you for your help. So then the difference between AC and A'C' (which I had asked in comments) should be : w(a/b+b/a) is that right? $\endgroup$ – Kumar Jul 12 at 16:58
  • $\begingroup$ @Kumar If you mean $AC-A'C'$ then it's correct $\endgroup$ – Akash Karnatak Jul 13 at 2:16
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Draw a perpendicular from $A'$ to $AC$. Let $X$ be the point of intersection of the perpendicular and $AC$. $A'AX$ is a right triangle, similar to $ABC$. $A'X=W$. From similarity of triangles, $AX:W=AB:BC$ or $AX=\frac{AB \cdot W}{BC}$. Now use Pythagorean theorem to find $AA'$. Similarly, construct triangle $CC'Y$ and obtain that $CY=\frac{BC \cdot W}{AB}$.

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