In triangle ABC prove two lines are perpendicular. 
In triangle ABC , D and E are the feet of heights from C and B. Prove the radius OA of circumscribed circle is perpendicular to DE.
Note: I used this as an statement to solve another problem in this site. 
 A: The complete solution:
This is actually an easy one. I give you the steps you need to follow to prove what you want.

Let $\measuredangle ABC=\beta$.
Since $BDC$ is a right-angled triangle, $\measuredangle DCB=90^0-\beta$.
Since $\measuredangle AOC$ is the subtended angle at the center of circle, $\measuredangle AOC=2\beta$.
Since $AOC$ is an isosceles triangle, $\measuredangle AOC=90^0-\beta$.
Consider the right-angled triangle $BEC$. Its circumcircle has its center at the midpoint of $BC$ and its radius is equal to $\frac{BC}{2}$. Now, consider the right-angled triangle $BDC$. Its circumcircle also has its center at the midpoint of $BC$ and its radius is equal to $\frac{BC}{2}$. This proves that both triangles have the same circumcircle. Therefore, $BDEC$ is a cyclic quadrilateral. Since all angles (in our case $\measuredangle DCB$ and $\measuredangle DEB$ inscribed in a circle (in our case the circumcircle of the quadrilateral $BDEC$) and subtended by the same chord (in our case $BD$) are equal, i.e. $\measuredangle DEB=90^0-\beta$.
Since $\measuredangle BEA$ is a right angle, $\measuredangle AED=90^0-\measuredangle DEB=\beta$.
Name the intersection point of the two lines $AO$ and $DE$ as $F$. Now, consider the triangle $AEF$. We know that two of its angles, i.e. $\measuredangle AEF=\beta$ and $\measuredangle FAE=90^0-\beta$. Therefore, the remaining angle $\measuredangle EFA$is equal to $90^0$ and this proves that $OA$ is perpendicular to $DE$.
