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Finding distance between the cross point of the diagonals(O) of a right-angled trapezoid and point lying on the right-angled side(BC), if the size of the small base(DC) is 4 and the size of the large base is 8. Here is drawing:

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I'm searching the size of OX(the red line). The first thing that came into my mind is that OX is perpendicular to BC from Shortest line segment theorem, whoever I don't know how to continue.

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  • $\begingroup$ We can see $\dfrac{4}{8}=\dfrac{DO}{OB}$, now try to figure out $\dfrac{OX}{DC}=\dfrac{OB}{DB}=?$ and then find $OX$ $\endgroup$ – Amir Naseri Mar 31 '16 at 16:02
  • $\begingroup$ I got the answer ($$\frac{8}{3}$$ which seems to be right, whoever I don't get from where did you got that $$\frac{4}{8}=\frac{DO}{OB}$$ $\endgroup$ – Planet_Earth Mar 31 '16 at 18:40
  • $\begingroup$ It comes from $\Delta OAB$ and $\Delta OCD$ similarity, and shows $OX=\dfrac83$ as well. $\endgroup$ – Amir Naseri Mar 31 '16 at 19:13
  • $\begingroup$ Thank you, if you give those comments as answer, I will mark it as "best answer" $\endgroup$ – Planet_Earth Mar 31 '16 at 19:43
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Since $AB$ and $DC$ are parallel lines, we could observe $\Delta OAB$ and $\Delta OCD$ are similar, therefore:

$$\frac{OC}{OA}=\frac{DC}{AB}=\frac48$$ Now consider $\Delta OCX$ and $\Delta ACB$ similarity: $$\frac{OX}{AB}=\frac{OC}{AC}=\frac{OC}{OC+OA}=\frac{4}{4+8}=\frac13\Rightarrow OX=\frac{AB}{3}=\frac83 \quad\checkmark$$

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