Three Altitudes of a triangle are concurrent

Question

I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.

I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.

Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.

Answer 1

Since $AEFC$ is cyclic, $\angle FEB =\angle ACB$.

Since $BEDF$ is cyclic, $\angle FEB=\angle BDF$.

Thus $\angle GDF+\angle ACB=180^\circ$.

Answer 2

The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$

Answer 3

Expanded answer:

Take the diagram as shown, with $AF \perp BC$ and $CE \perp AB$ but consider the point $G$ to be merely the extension of $BD$ to meet $AC$ - angle unknown.

Now we find that $E$ and $F$ are points on the circle with $AC$ as diameter, giving $AEFC$ as cyclic, so we can see that the inscribed angle from chord $AF$ means that the internal angle $\angle ACF=\angle FEB$, the external angle at $E$.

Similarly $E$ and $F$ are points on the circle with $BD$ as diameter, giving $DEBF$ cyclic and the inscribed angle from chord $BF$ gives $\angle FEB=\angle FDB$.

Then since $\angle ACF=\angle FDB$ we have that $\angle ACF+\angle FDG = 180^\circ$ and completing the angles in quadrilateral $GDFC$ we see that $\angle CGD = 90^\circ$ and $BG$ is the third altitude.