# Finding distance between the cross point of the diagonals of a right-angled trapezoid and point lying on the right-angled side

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:

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.

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

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: $$\frac{OX}{AB}=\frac{OC}{AC}=\frac{OC}{OC+OA}=\frac{4}{4+8}=\frac13\Rightarrow OX=\frac{AB}{3}=\frac83 \quad\checkmark$$