Adding Points to the Picture

A problem about five points — what more could one ask for? Paradoxically it cannot be done without adding points. The equal angles ought to suggest something.

The equal angles can be used by intersecting suitable pairs of extended sides, producing isosceles triangles. The perpendicular bisectors of sides suddenly have a completely different meaning.

Point $K$ on the median looks strange. In any case, we need to use the fact that it is a median, i.e. that $M$ is a midpoint — ideally in a way that helps us better understand the equality $AB=CK$.

The key is to reflect $A$ through $M$ to $A'$, which gives strong meaning to $M$ since the equality $AB=CK$ suddenly becomes $A'C=CK$.

In the problem we have $H$, the midpoint of $BC$, and somehow line $MH$ also plays a role. Well, what are we going to do with that orthocenter 🙂?

If $H'$ is the image of $H$ under the point reflection through $M$, then we know it lands on the circumcircle of $ABC$ at the point opposite $A$. Have some cyclic quadrilaterals appeared?

The angle between the midline and a diagonal of the quadrilateral is not something I see every day. The key will be to focus on the midpoints themselves. What about adding one or more further midpoints?

A good midpoint to add is the midpoint of $AB$ or $CD$. Thanks to the parallelism coming from the midlines we can neatly rephrase the condition from the statement. Then we finish it off.

Orthocentres are complicated points if we draw all the altitudes. Discard such a drawing immediately. Their images under suitable reflections give a better understanding of the situation. It also helps to rephrase what we are actually proving, so that after reflecting the orthocenters we can still state the claim clearly.

First, it suffices to show that the segment connecting the orthocenters has the same length as $CD$ (then we have a parallelogram). This realization gives us confidence that when we reflect these orthocenters over an axis, we still have a good statement to prove, namely the equality of two chords of the circle.

We have an orthocenter. It may not be immediately obvious what to do, but what about reflecting it over something?

Let $H'$ be the image of $H$ under reflection over $AB$. It lands on the circumcircle of $ABCD$. Now the diagram contains everything we need and it only remains to connect a few facts.

We need to use the curious condition $\angle BMD=90^\circ$. Point $M$ is the midpoint of a segment, which already suggests something, and the right angle only reinforces the hint. In any case it pays to use colored pencils to mark equal-length segments.

The key is to reflect, say, $D$ through $M$. After marking all the equal segments arising from the resulting parallelogram and the isosceles triangle, something should catch your eye.

It pays to draw segment $AD$ horizontally (unconventionally so), giving the best diagram (redraw right now if you haven't). Even in the new diagram it may not be clear what should happen. The midpoint of $AD$ does not look easy to handle. Perhaps it needs some further midpoints to cooperate with. Ideally ones that are also related to points $K$ and $L$.

The key is the remaining midpoints of the sides of triangle $ABP$. Then magic happens and we are just a few easy arguments from the solution.

Try to shift point $P$ so that the condition on the sum of angles makes sense.

The key point is the image $P'$ of point $P$ under the translation in the direction of $AC$ (or $BC$). Then we have a cyclic quadrilateral $CP'DP$ as well as plenty of parallelograms.

The angle sum condition is key. We need to introduce a point that exploits the angle-sum condition and makes it a little more meaningful.

The key point is the reflection of $C$ in $BD$, or analogously the reflection of $E$ in $DF$. It is the same point. And it is all we need. Do not forget to use the inequality part of the condition $CD=DE \neq AD$. Without that, the problem does not hold.

The angles imply parallel lines. They hint at the right move forward.

The point to introduce is the intersection point $X$ of $BC$ and $AD$, as $PCXD$ is now a parallelogram. The problem can now be solved with the right view on what we have.

The situation is fairly involved and one cannot discover much that is interesting at first glance. However, there is a point $H$ in it with which we can do all sorts of interesting things. Which of those is suitable here?

Reflect $H$ over $BC$ to obtain $H'$ lying on the circumcircle of $ABC$. This means $OA = OH'$. The midpoint of $OH$ also becomes more tractable.

The midpoint of $AP$ is a strange point that does not quite fit. Unless we use it for a point reflection to transfer one of the angles whose equality we are proving.

It makes sense to reflect $Q$ through $M$ to transfer angle $PQA$; it suddenly equals $PQ'A$. The statement to be proved will then, after a brief angle chase, tell us what it suffices to show.

The condition on the positions of $K$ and $L$ is interesting. Rewriting it in ratio form suggests that some homothety is at play. What about reflecting something further in it?

Consider the homothety mapping $AB$ to $DC$; under it, $K$ maps to $L$. If we map $P$ to $P'$ under it, then points $K$, $L$, $P$, $Q$, $P'$ all lie on a line. The problem turns into a nice angle-chasing exercise.

Draw only one point, say $K$, and try to better understand the strange angle condition by which it is defined. Adding further points will start to make sense only once you discover something.

In the diagram with only point $K$ one can angle-chase $\angle BKC=\angle KDC$. Line $AB$ is therefore tangent to the circumcircle of $KCD$. This is reminiscent of the power of a point — one just needs to add the point from which we compute it. Once this is clear, we can connect the remaining points $L$, $M$, $N$.

We have many midpoints. Do we reflect or do we introduce even more? Here the latter makes more sense.

The key point to introduce is the midpoint of $XY$, as it will create many midlines. And they will give a clear path of what to use to prove the concyclicity.

Point $Q$ is defined in a fairly arbitrary way. There is nothing for it but to add some further point that brings meaningful structure. The trapezoid itself gives a fairly good hint.

Reflect $Q$ under the homothety centred at $P$ that maps $AB$ to $CD$. Suddenly the definition of $Q$ changes into something meaningful and out of nowhere we have everything we need — we just put it together.

We have $FB = FD$, but it is not clear how to use this. It gives us equal angles $\angle FBD$ and $\angle FDB$. The second one in particular looks stranded. Try intersecting something with something to transfer these angles somewhere.

The key is to extend segment $FD$ to a chord of $\Gamma$, and similarly extend $GE$ on the other side. Suddenly we obtain isosceles triangles that interact well with the condition $AD = AE$. No further points are needed — it becomes an angle-chasing exercise.

We are almost given a parallelogram with those parallel lines.

Let $Z$ be the intersection point of the parallel lines from the statement — then using the homothety at $A$ with coefficient $2$ we have that $\angle KIL = \angle BZC$. We do not have those ugly midpoints anymore as we are proving just $\angle BZC + \angle YPX = 180^\circ$.

The key to finish the problem is to find cyclic quadrilaterals. But at least we do not need to introduce more points.

The most problematic point is $M$: as a midpoint it has no nice angles around it. It needs to be given better meaning. The statement to be proved also hints at what we want to do.

The key is to reflect, say, $R$ through $M$. Then it pays to color equal-length segments; the goal is to color another pair. We are not far from the desired end.

Point $S$ is the midpoint of some segment and this makes angle chasing infeasible. Still, something can be angle-chased, which may suggest how to use midpoint $S$ to define a new point.

The key is to discover $AR_0 \parallel RJ$. This gives a fairly clear hint: the image of $J$ under the point reflection through $S$ lands on $AR_0$. We wrap it up from here — we have everything needed for angle chasing.

The problem almost invites us to define the second intersection of circles $k_1$ and $k_2$. This naturally interweaves them, and we can also make good use of their diameters.

If $T$ is the second intersection of $k_1$ and $k_2$, then the diameters give us right angles, from which $T$, $X$, $Y$ are collinear. The proof that $H$ also lies on this line can be rephrased as a perpendicularity. Have we accidentally eliminated $X$ and $Y$?

Points $Q$ and $P$ are midpoints of $AN$ and $AM$ respectively. This fact is hard to use on its own. The key is to define some further midpoints. At best these midpoints will help us eliminate the not-so-natural-looking points $M$ and $N$.

Define midpoints $J$ and $K$ of $AC$ and $AB$ respectively. Midlines give us $BM \parallel KP$ and $CN \parallel QJ$. So what is required is to show that the angle between segments $QJ$ and $KP$ is $180^\circ - \alpha$. We seem to have made no progress, as angles at medians are generally ugly. However, there are many similar triangles in the configuration.

Points $O_1$ and $O_2$ look genuinely artificial. Indeed they are. Both can be eliminated by defining a new point and rephrasing the parallelism into something else. Even without these strange points, more holds in the configuration than the problem states, and it is not very easy to see. It pays to guess it.

Let $T$ be the second intersection of our two circles with centres $O_1$ and $O_2$. We want to show that $AT$ is perpendicular to $I_1I_2$. This is still not free. It helps to guess and prove that $BCI_2I_1$ is a cyclic quadrilateral.

The problem is a little awkward to draw. Do not be intimidated. The non-standard angle condition looks like it could mean something nice. It does — one just needs to intersect something. As for angle bisectors, do not be afraid of them; two often imply a third and here there are plenty.

Let $E$ lie on the bisectors at $A$ and $B$, $F$ on the bisectors at $B$ and $C$, $G$ on the bisectors at $C$ and $D$, and $H$ on the bisectors at $D$ and $A$. All points $E$, $F$, $G$, $H$ are incenters or excenters of some triangles thanks to the angle bisectors. In any case, points $P$, $F$, $H$ and $Q$, $E$, $G$ are collinear, and these lines are the angle bisectors of $\angle CPB$ and $\angle CQD$. We are now very close.

It is obvious that something is missing from the problem. It therefore pays to first get a better picture of the situation by rephrasing the statement as an equality of some other angles, ideally ones expressible in terms of the basic angles of $ABC$. This may inspire an addition. Let us not forget that the problem still contains the well-known indicator of potentially good additions: a right angle.

The statement is equivalent to $\angle PCQ=\angle OCB$, which is equivalent to $\angle BAC=\angle CQP$. So if we reflect $Q$ through $P$ to $Q'$, it suffices to prove concyclicity. That too is not obvious, but it can be finished at this point.

When we do not know what to do, chase angles for a bit. This should let us find a simple nice property in the diagram that may inspire what to add.

The key is to discover $BC \parallel DE$. The diagram contains the intersection of $CE$ and $BD$. This may inspire adding the image of $A$ under the homothety mapping $BC$ to $DE$. Suddenly everything becomes beautiful.

The goal is to add something that clarifies the strange angle condition for $P_1$ and $Q_1$. However the problem has too many points and various things that look promising can be added without much help — for instance the intersection of $P_1P$ with $BC$ or of $PQ$ with $AB$ (try it). The trouble with these intersections is that while they do give a cyclic quadrilateral, it provides no further useful angles to transfer. It is therefore better to intersect some lines with a circle in a way that introduces one more angle into the equality $\angle PP_1A=\angle CBA$.

The key magic point is the second intersection of line $BB_1$ with the circumcircle of $ABC$, call it $C_2$. Thanks to it we get $\angle CBA=\angle CC_2A$, which magically gives the cyclic quadrilateral $C_1C_2P_1A$. Analogously we define $B_2$. And believe it or not, it is now only a moderately hard angle-chasing problem.

Nothing meaningful can be seen in the configuration. Something is hidden there — but it is not easy to see exactly what. One option is to try to do something meaningful with lines $BX$ and $CX$. Points $K$ and $L$ lie on them and we do not see how to use that. One possibility is to intersect them with something meaningful.

The key points are the intersections $E$ and $F$ of lines $BX$ and $CX$ with the circle on diameter $BC$. Even those alone do not suffice — but the next point is quite natural: the intersection $G$ of lines $BF$ and $CE$. It clearly lies on line $AXD$ and $X$ is the orthocenter of triangle $GBC$. Although it may not seem so, the problem can now be finished — by combining power of a point, inscribed angles and cyclic quadrilaterals. Even at this stage it is not an easy problem.

When we have a sum of areas, it is better to interpret it nicely. But these triangles overlap awkwardly. So we better 'move' these areas elsewhere.

The trick is that if $X'$ is any point on the line through $X$ parallel to $BC$, then $A_{BCX}=A_{BCX'}$. What is this hinting us at?