Prove a collinearity equivalence (Euclidean geometry)

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.

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.

• Fantastic! It never occurred to me that Pascal's theorem is an "if and only if", as I have only seen the forwards direction used until now. Commented Jan 8 at 23:02
• Yes, the "conic" version of Pascal theorem is "if and only if". Commented Jan 9 at 0:21

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 $$\angle OBX = 90^\circ - \angle CAB$$.

• Amazingly elegant and well-written solution! Commented Dec 29, 2023 at 4:20
• I'm a bit confused on how (♡) supports the equations above it, because a = b doesn't imply that a = b = 1. Commented Jan 8 at 23:07
• @user1050148 The point is that if $a=1$ and $a=b$ then $b=1$. Similarly, if $b=1$ and $a=b$ then $a=1$. Commented Jan 9 at 0:19