A triangle $OAB$ is right angled at $O;$squares $OALM$ and $OBPQ$ are constructed on the sides $OA$ and $OB$ externally A triangle $OAB$ is right angled at $O;$squares $OALM$ and $OBPQ$ are constructed on the sides $OA$ and $OB$ externally.Show that the lines $AP$ and $BL$ intersect on the altitude through $'O'$.

This question is asked in the vector section of my book.I tried to solve but in vain.I dont know what is the proper way to prove this question.Please help me.
 A: Let the intersection of $AP$ and $BL$ be $O'$.
Draw square $ABXY$ external to side $AB$, connect $OX,OY$.
Draw perpendicular lines from $A$ to $BL$ and from $B$ to $AP$, let the intersection be $Z$ (i.e. essentially we just make triangle $ABZ$ that $AZ\perp BL$ and $BZ\perp AP$)
Now first since $\triangle ABL\cong \triangle AYO$ we have the angle between $OY$ and $BL$ is the same as the angle between $AB$ and $AY$, which is ninty degrees. Hence $OY\perp BL$. Similarly $OX\perp AP$ as well.
Now we have $OY\parallel AZ$ and $OX\parallel BZ$ and $AB\parallel XY$.  From here we have $\angle ZAB=\angle OYX$ and $\angle ZBA=\angle OXY$. Furthermore since $AB=XY$ we have $\triangle ZAB\cong \triangle OYX$.
From this we know $ZOYA$ is a parallelogram and $ZO\parallel AY$ and hence $ZO\perp AB$. Furthermore, since $O'$ is the intersection of two altitudes of triangle $ZAB$, we have $O'$ is the orthocenter of $\triangle ZAB$. Hence line $ZO$ which is the other altitude must pass through $O'$. From here we know $OO'$ is on the same line as $ZO$ and is perpendicular to $AB$. QED.
