Space Geometry: lines in a plane 
If $d$ and $d'$ are two intersecting lines in a plane $P$, and $D$ is a line orthogonal to both $d$ and $d'$, then any line $\delta$ in $P$ is orthogonal to $D$ as well. 

How could this be proven using only the tools of Euclidean geometry?
I tried to do that by contradiction, but this seems to be an unfruitful approach. 
Thank you.
 A: We will use the following defintion:   
Definition. A line $l$ is orthogonal to a plane $\Pi$ if it is orthogonal to every line in the plane, which passes through the intersection of the line $l$ and the plane $\Pi$.   
So, with this definition, you are asking for a proof of the following statement:   
Theorem. Every line that is orthogonal to two intersecting lines at their intersection, is orthogonal to the plane that is defined by the intersecting lines. 
Remember that a plane can be uniquely determined by two (distinct) intersecting lines. We will prove the theorem using a direct proof. The proof goes like this:    
Proof. Let $AB$ be a line orthogonal to the lines $BC$ and $BD$ at the point $B$ of their intersection. We will prove that the line $AB$ is orthogonal to the plane $\Pi$ which is defined by the lines $BC$ and $BD$. By definition, it suffices to show that $AB$ is orthogonal to every line in $\Pi$ which passes through $B$. Let $BE$ be such a line. 

Draw the line $CD$, which intersects  $BE$  (suppose) at the point $E$. Now, extend the line $AB$ and on this extension take a point $A'$ such that $BA'=AB$. Draw the lines $AC,AE,AD,A'C,A'E$ and $A'D$.
It is true that $CA=CA'$ because $CB$ is orthogonal to $AA'$ at $B$ which is the middle point of the segment $AA'$. Similarly, it is true that $DA=DA'$. The triangles $ACD$ and $A'CD$ are congruent because they have corresponding sides equal in length ($CD$ is a common side, and $CA=CA', \; DA=DA'$ as we said above) Thus, it is true that $\hat{D}_{1}=\hat{D}_{2}$.
On the other side, the triangles $AED$ and $A'ED$ are congruent because


*

*$DA=DA'$ as we said above,

*$DE$ is a common side,

*$\hat{D}_{1}=\hat{D}_{2}$  


Thus, it is true that $EA=EA'$.
Now, because of that, the triangle $AEA'$ is isosceles, which means that the median of $EB$ is orthogonal to the base $AA'$ at $AA'$ 's middle point $B$. Therefore, the line $AB$ is orthogonal to $BE$ which is what we wanted to prove.  $\square$
I hope that I helped you!
