Here is the orthologic triangles theorem:
$ABC$ and $A'B'C'$ are triangles in the same plane. If the perpendiculars from $A$, $B$ and $C$ to $B'C'$, $C'A'$ and $A'B'$ respectively are concurrent, prove that so are the perpendiculars from $A'$, $B'$ and $C'$ to $BC$, $CA$ and $AB$ respectively.
In my book two solutions have been given using coordinate geometry and another using vectors. But I was wondering if you can solve this using pure geometry. Any help would be appreciated.
Recall well-known lemma:
Lemma 1. Let $D,E,F$ be points on lines $BC, CA, AB$. Then lines perpendicular to $BC, CA, AB$ through $D,E,F$ are concurrent if and only if $BD^2+CE^2+AF^2=DC^2+EA^2+FB^2$.
Recall another known lemma:
Lemma 2. For any points $A,B,C,D$ the following equivalence holds: $AC \perp BD \iff AB^2+CD^2=BC^2+DA^2$.
We can easily generalize lemma 1 using lemma 2 to the following
Lemma 3. Let $A,B,C,D,E,F$ be any points. Then lines perpendicular to $BC, CA, AB$ through $D,E,F$ are concurrent if and only if $BD^2+CE^2+AF^2=DC^2+EA^2+FB^2$.
Now we are ready to prove the theorem.
First, we put $(A,B,C,D,E,F):=(A',B',C',A,B,C)$ in lemma 3 and we get the equality $$B'A^2+C'B^2+A'C^2=AC'^2+BA'^2+CB'^2.$$ Now we put $(A,B,C,D,E,F)=(A,B,C,A',B',C')$ in lemma 3 and conclude that the equality above implies that perpendiculars to $BC, CA, AB$ through $A',B',C'$ are concurrent. $\square$
There is a rotation and scaling (without translation) with respect to common orthocenter of ABC and A'B'C'. Thus the two triangles are duals, or, alternately belong to a set of dialation/rotation transforms.
So its proof is easier ( I believe) with complex number sets.
