Today, we had a lecture on Euler line and it's various generalizations/applications. One of the tasks we were given as homework is the following one:
Let acute $\triangle ABC$ have altitude $\overline{AD}$. Let $P$ and $Q$ be midpoints of sides $\overline{AB}$ and $\overline{AC}$, respectively; and let $X$ and $Y$ be the projections of $D$ onto sides $\overline{AB}$ and $\overline{AC}$, respectively. Lines $\overleftrightarrow{PY}$ and $\overleftrightarrow{QX}$ intersect at point $Z$. Prove that circumcenter $O$ is collinear with $Z$ and $D$.
Edit by @Blue. The original problem statement had a weaker condition on $P$ and $Q$, namely: "Let $P$ and $Q$ be such points, that $|AP|=|PB|$ and $|AQ|=|QC|$." However, the result with this condition is not true in general. I strengthened the condition to make $P$ and $Q$ midpoints, but the original source might have intended something else (otherwise, why not just call $P$ and $Q$ midpoints from the get-go?).
This is supposedly only something like a generalization of Euler's line and should be possible to prove using methods similar to those used to prove that Euler's line always exists, yet I'm having difficulty in proving this theorem, as I fail to notice any helpful analogy. (I managed to find an analytic proof of this theorem, but it's over four pages long and we were not supposed to use such techniques.)
Image by @Blue. (This assumes the midpoint condition.)
3 Answers
$\begingroup$Below is a proof of a slight generalization of "Blue's version" of the problem. The slight generalization is that $\overline{PQ}$ is merely assumed to be parallel to $\overline{BC}$; the conclusion is that $D$, $E$, and $Z$ in the figure are collinear.
To recap, we are assuming that:
- $\triangle ABC$ is a triangle with an altitude $\overline{AD}$
- $X = D_b$, where "$D_b$" denotes the projection of $D$ onto $\overline {AB}$
- $Y=D_c$, where "$D_c$" denotes the projection of $D$ onto $\overline{AC}$
- $\overleftrightarrow{PQ}$ is parallel to $\overline{BC}$, with $P$ on $\overline{AX}$ and $Q$ on $\overline{AY}$
- $Z$ is the intersection of $\overline{PY}$ and $\overline{QX}$
Draw the line through $P$ perpendicular to $\overline{AB}$ and the line through $Q$ perpendicular to $\overline{AC}$, and let $E$ be the intersection of those two lines. (In other words, $E_b = P$ and $E_c = Q$.)
The points on line $\overleftrightarrow{DE}$ are the points which can be written in the form $(1-\lambda)D + \lambda E$ (with $\lambda \in \mathbb{R}$). For all $\lambda$, the point $K = (1-\lambda)D + \lambda E$ satisfies \begin{align*} K_b &= (1-\lambda)D_b + \lambda E_b = (1-\lambda)X + \lambda P \\ K_c &= (1-\lambda)D_c + \lambda E_c = (1-\lambda)Y + \lambda Q. \end{align*} The "converse" is also true, in the following sense: For any point $K$ (not necessarily on $\overleftrightarrow{DE}$), let $\lambda_b(K)$ be the unique number $\lambda_b$ such that $K_b = (1-\lambda_b)X + \lambda_b P$, and let $\lambda_c(K)$ be the unique number $\lambda_c$ such that $K_c = (1-\lambda_c)Y + \lambda_c Q$; then $\overleftrightarrow{DE}$ is the locus of points $K$ for which $\lambda_b(K) = \lambda_c(K)$.
(You could think of $(\lambda_b, \lambda_c)$ as a bizarre oblique coordinate system for the plane.)
Oh yeah, so what were we trying to prove again?
Claim: $Z$ lies on $\overleftrightarrow{DE}$.
Special case (corollary): In the case where $P$ is midpoint of $\overline{AB}$ and $Q$ is the midpoint of $\overline{AC}$, the point $E$ is in fact the circumcenter $O$. Hence the conclusion is taht $Z$ must lie on line $\overleftrightarrow{DO}$.
Proof of the claim:
Use the right angles at $X$, $D$, and $Y$ to show that $\angle ADX = \angle B = \beta$ and $\angle ADY = \angle C = \gamma$.
Since $\angle AXD$ and $\angle AYD$ add up to $180^{\circ}$, the quadrilateral $AXDY$ is a cyclic quadrilateral. This implies that $\angle AYX = \angle ADX$ and $\angle AXY = ADY$.
We now know the angles at all four vertices of the quadrilateral $XYQP$: $\angle X = \gamma$ and $\angle Y = \beta$ from the above, and $\angle Q = 180^{\circ}-\gamma$ and $\angle P = 180^{\circ}-\beta$ since $\overline{PQ}$ is parallel to $\overline{BC}$.
These angles show that $XYQP$ is a cyclic quadrilateral. In particular, this means that $\triangle PZX$ and $\triangle QZY$ are similar. The similar triangles imply that $\lambda_b(Z) = \lambda_c(Z)$.
From the earlier discussion, we conclude that $Z$ lies on $\overleftrightarrow{DE}$.
(Remark: The center of the circle passing through $X$, $Y$, $Q$, and $P$ also lies on $\overleftrightarrow{DE}$, since it has $\lambda_b = \lambda_c = 1/2$.)
But wait, what does it have to do with the Euler line?
Uhhhh... nothing really, as far as I can tell. In my mind the Euler line has to do with that one neat trick of using a dilation by a factor of $-2$ centered at the centroid. This problem doesn't seem to involve that idea at all. In my proof here, the outline was that $\overleftrightarrow{DE}$ is characterized by "$\lambda_b = \lambda_c$", and $Z$ has that property. Of course, I don't know what proof of the existence of the Euler line is the one you have in mind, but I personally don't see any helpful analogy.
$\endgroup$ 7 $\begingroup$Here's a coordinate proof of the generalization I made to @echinodermata's generalization. (I know: Coordinates are not the preferred technique. However, the trigonometric aspects of the results may suggest a route to a purely geometric solution.)
Suppose $\overleftrightarrow{OR}$ bisects both $\angle POQ$ and $\angle DOE$. Let $D^\prime$ and $E^\prime$ be projections onto $\overleftrightarrow{OP}$ of $D$ and $E$, respectively; likewise, let $D^{\prime\prime}$ and $E^{\prime\prime}$ be projections onto $\overleftrightarrow{OQ}$. If $\overleftrightarrow{D^\prime E^{\prime\prime}}$ and $\overleftrightarrow{D^{\prime\prime}E^\prime}$ meet at $F$, then $D$, $E$, $F$ are collinear. (If the lines are parallel, we can take $F$ to be the "point at infinity", and the result holds.)
Proof. Define $$\theta := \angle POR = \angle QOR \qquad \phi := \angle DOR = \angle EOR \qquad d := |\overline{OD}| \qquad e := |\overline{OE}|$$ (where, throughout, we might as well allow angles and lengths to be negative) so that $$\frac{|\overline{OD^\prime}|}{d} = \cos(\theta-\phi) = \frac{|\overline{OE^{\prime\prime}} |}{e} \qquad \frac{|\overline{OD^{\prime\prime}}|}{d} = \cos(\theta+\phi) = \frac{|\overline{OE^{\prime}} |}{e}$$
(For those seeking a geometric solution, it is perhaps worth noting that $\triangle D^\prime O E^{\prime\prime} \sim \triangle D^{\prime\prime} O E^\prime$; therefore also, $\triangle D^\prime F E^\prime \sim \triangle D^{\prime\prime} F E^{\prime\prime}$.)
Coordinatizing, with $O$ at the origin, and $R$ along the positive $x$-axis, and writing $\operatorname{cis}\theta$ for $(\cos\theta,\sin\theta)$, we have $$D = d\operatorname{cis}(\phantom{-}\phi) \qquad D^\prime = d\cos(\theta-\phi)\operatorname{cis}\theta \qquad D^{\prime\prime} = d\cos(\theta+\phi)\operatorname{cis}(-\theta)$$ $$E = e\operatorname{cis}(-\phi) \qquad E^\prime = e\cos(\theta+\phi)\operatorname{cis}\theta \qquad E^{\prime\prime} = e\cos(\theta-\phi)\operatorname{cis}(-\theta)$$
(We see here that we should disallow $\cos(\theta\pm\phi)=0$, so that $\overleftrightarrow{D^\prime E^{\prime\prime}}$ and $\overleftrightarrow{D^{\prime\prime} E^\prime}$ are defined.) The relevant line equations are straightforward to derive: $$\begin{align} \overleftrightarrow{DE}&: \quad x (d+e) \sin\phi - y (d-e)\cos\phi = 2 d e \sin\phi \cos\phi \\[6pt] \overleftrightarrow{D^\prime E^{\prime\prime}}&: \quad x (d+e) \sin\theta - y (d-e) \cos\theta = 2 d e \cos(\theta-\phi) \sin\theta\cos\theta \\[6pt] \overleftrightarrow{D^{\prime\prime} E^\prime}&: \quad x (d+e) \sin\theta + y (d-e) \cos\theta = 2 d e \cos(\theta + \phi) \sin\theta\cos\theta \end{align}$$
If $d\pm e = 0$, the last two lines are parallel, giving the "point at infinity" for $F$. Otherwise, we can solve to get $$F = 2 d e\,\left( \frac{\cos^2\theta \cos\phi}{d + e}, - \frac{ \sin^2\theta \sin\phi}{ d - e }\right)$$ which evidently satisfies the equation for $\overleftrightarrow{DE}$. $\square$
To tie this back to the original question, we note that, in $\triangle ABC$, the altitude from $A$, and the line joining $A$ to the circumcenter, are reflections of each other in the bisector of $\angle A$. The points $P$, $Q$, $X$, $Y$ there are projections of the altitude-foot and the circumcenter, just as here, points $D^\prime$, $D^{\prime\prime}$, $E^\prime$, $E^{\prime\prime}$ are projections of $D$ and $E$. And, of course, the question's $Z$ corresponds to $F$ here.
$\endgroup$ $\begingroup$Synthetic solution: Let M,P,N be midpoint of OD,OA,AD. Because P,N is center of AEOF,AHDG and MN,MP perp to GH,EF so M is center of EFGH. HM, GM cuts (M) at K, L then by Pascal M,O,Z collinear so is D,O,Z
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