Prove a collinearity equivalence (Euclidean geometry)

Question

In an acute-angled triangle $ABC$, let $AD$ and $BE$ be altitudes and let $AP$ and $BQ$ be internal angle bisectors. Denote by $I$ and $O$ the incentre and the circumcentre of $ABC$, respectively. Prove that $D$, $E$, and $I$ are collinear iff $P$, $Q$, and $O$ are collinear.

Although this problem can be readily attacked by cartesian coordinates, and maybe even complex numbers, I am searching for a synthetic or trigonometric (sine rule and cosine rule) solution to this problem so I can present it to a class.

I couldn't get the Simpson line setup to work here, and I hopelessly tried using Menelaus on $DPCAEID$ (I assumed DEI collinear), but after turning the ratios $\frac{DP}{DI},\frac{AC}{PC},\frac{EI}{AE}$ into ratios of sines using the sine law, I ended up with 1 = 1.

Are there any other synthetic theorems/tools for collinearity that can be used to attack this?

Hints or a full solution would both be appreciated.

Answer 1

Let me post yet another solution.

Let $\omega$ be the circumcircle of $ABC$. Let $AI$ and $BI$ intersect $\omega$ at $M\neq A$ and $N\neq B$, respectively. Let $MD$, $NE$ intersect at $T$. Note that by Pascal theorem for the hexagon $ACBNTM$ the points $D,I,E$ are collinear if and only if $A,C,B,N,T,M$ are conconic. Since the conic passing through $A,B,C,M,N$ is $\omega$, this is equivalent to saying that $T$ lies on $\omega$.

Similarly, let $AO$, $BO$ intersect $\omega$ at $U \neq A$ and $V \neq B$. Let $UP$, $VQ$ intersect at $T'$. Then collinearity of $P,O,Q$ is equivalent, by Pascal theorem for $ACBVT'U$, to conconicity of $A,C,B,V,T',U$, and again this is the same as saying that $T'$ lies on $\omega$.

Now, suppose that $D,I,E$ are collinear. Then, as discussed above, $T$ lies on $\omega$. It is well-known (and easy to prove) that $NE\cdot NT = NA^2 = NQ \cdot NB$, hence $B,Q,E,T$ are concyclic. Since $QEB=90^\circ$, it follows that $\angle QTB = 90^\circ$. But $BV$ is a diameter of $\omega$, hence $\angle VTB=90^\circ$. This shows that $V,Q,T$ are collinear. Analogously, $U,P,T$ are collinear. This shows that $T=T'$. As such, $T'$ lies on $\omega$. Therefore $P,O,Q$ are collinear.

We prove the other direction now. Suppose that $P,O,Q$ are collinear. Then $T'$ lies on $\omega$. Note that $\angle VT'B = 90^\circ = \angle QEB$, hence $QET'B$ is cyclic. Since $\angle ET'Q = \angle EBQ = \angle NBV = \angle NT'V$, it follows that $N,E,T'$ are collinear. Similarly, $M,D,T'$ are collinear. This shows that $T'=T$, hence $T$ lies on $\omega$ and, as proved in the beginning, $D,I,E$ are collinear.

Answer 2

Let $AO$ and $BO$ cut $BC$ and $AC$ at $X$ and $Y$, respectively.

By Menelaos for triangle $ACP$, the points $D,I,E$ are collinear if and only if $$\frac{AE}{EC}\cdot\frac{CD}{DP}\cdot\frac{PI}{IA}=1.$$

By Menelaos for triangle $AXC$, the points $P,O,Q$ are collinear if and only if $$\frac{AQ}{QC}\cdot \frac{CP}{PX}\cdot \frac{XO}{OA}=1.$$

Therefore we will be done once we prove that

$$\frac{AE}{EC}\cdot\frac{CD}{DP}\cdot\frac{PI}{IA} = \frac{AQ}{QC}\cdot \frac{CP}{PX}\cdot \frac{XO}{OA}. \tag{$\heartsuit$}$$

Angle bisector theorem yields the following: $$\frac{PI}{IA}=\frac{PB}{BA} = \frac{PC}{CA},\qquad \frac{AQ}{QC}=\frac{AB}{BC}, \quad \text{and} \quad PX=\frac{AX \cdot PD}{AD}.$$

Susbtituting this to $(\heartsuit)$ we obtain equivalently

$$\frac{AE}{EC} \cdot \frac{CD}{DP} \cdot \frac{PC}{CA} = \frac{AB}{BC} \cdot \frac{CP}{\frac{AX \cdot PD}{AD}} \cdot \frac{XO}{OA}.$$

Cancelling out common terms we obtain equivalently

$$\frac{AE}{EC}\cdot\frac{CD}{AC} = \frac{AB}{BC}\cdot \frac{AD}{AX} \cdot \frac{XO}{OA}. \tag{$\spadesuit$}$$

Note that $\triangle ACD \sim \triangle BCE$, hence $\dfrac{CD}{AC}=\dfrac{CE}{BC}$. Substituting this to $(\spadesuit)$ and cancelling out $EC$ and $BC$ we obtain equivalently

$$AE=AB\cdot \frac{AD}{AX}\cdot\frac{XO}{OA}.\tag{$\clubsuit$}$$

Note that $AE=AB\cos\angle CAB$, $AD=AX\sin\angle DXA$, and $\dfrac{XO}{OA}=\dfrac{XO}{OB}=\dfrac{\sin \angle OBX}{\sin\angle BXO}$. Substituting this to $(\clubsuit)$ and cancelling out whatever can be cancelled out we are left with proving that $$\cos \angle BAC = \sin \angle OBX.$$ This is clearly true since $\angle OBX = 90^\circ - \angle CAB$.