A geometry problem to prove three line segments intersect at the same point

Question

$D$ is a point within $\triangle ABC$ such that $\angle DBA = \angle DCA$. $I$ is midpoint of $BC$. $DE \perp AB$, $DF \perp AC$. $E,F$ are on $AB$, $AC$ respectively. $G$ is a point such that $AG \perp GD$, $AG$ and $IF$ meet at $H$.

Prove that $EF, BH, DG$ intersect at the same point.

Through software, apparently we have $\angle EIF = 2\angle EBD$, though I can't prove it easily (would have been trivial if $CDE$ and $BDF$ are both colinear.

Also what are the arsenals to prove three line segments meet at the same point? Perhaps Pascal's theorem?

Answer 1

Any theorem worth using once is worth using four times!

Observe that $A, D, E, F, G$ lie on a circle $Γ$ centered at the midpoint $J$ of $AD$. Let $BD, CD$ meet $Γ$ again at $K, L$. Let the tangents to $Γ$ at $E, K$ meet at $M$, and the tangents at $F, L$ meet at $N$.

Since

$$\begin{multline*}\frac{∠KME}{2} = \frac{180° - ∠EJK}{2} = ∠KDE - 90° = ∠KBE \\ = ∠FCL = ∠FDL - 90° = \frac{180° - ∠LJF}{2} = \frac{∠FNL}{2},\end{multline*}$$

we see that

• $B, E, K$ lie on a circle $Γ_1$ centered at $M$;
• $C, F, L$ lie on a circle $Γ_2$ centered at $N$;
• there’s a diameter $\ell$ of $Γ$ that reflects $E, K, M, Γ_1$ to $F, L, N, Γ_2$.

Let $EL, FK, \ell$ meet at $I'$. Using Pascal’s theorem twice on $EEFKKL$ and $EFFKLL$, we find that $I'$ lies on $MN$, so it is in fact the midpoint of $MN$. Using Pascal’s theorem on $AELDKF$, we find that $I'$ lies on $BC$. This is enough to establish that $I' = I$, the midpoint of $BC$.

Finally, the result follows using Pascal’s theorem on $AEFKDG$.