Isosceles triangle with the orthocenter on the base

Today we will revisit a rather old but underrated configuration which has certain beautiful properties. Our work shall revolve around the points $E$ on $AC$ and $F$ on $AB$ such that $AE=AF$ and points $E,H$ and $F$ lie on a straight line.

Shoutouts to raosicheng for suggesting an overwhleming majority of these problems and mathematical_arceus77 for posting most of them on AoPS.

Here is the picture we shall be referring to.

A summary of the most important lemmas from this configuration,

First, circle passes through the Queue point which has been questioned previously, most prominently in Shortlist 2005 G5.
- An immediate result is that points $L$ , $O$ and $O'$ – the circumcenter of $(AEF)$ are collinear since they all lie on the perpendicular bisector of segment $AQ_a$ .
- Hence, $Q_a$ is the center of spiral similarity mapping segment $EF$ to $BC$ . This then allows us to conclude that quadrilaterals $QFBG$ and $QECG$ are both cyclic.
- As a result it follows that points $G$ , $Q_a$ and $M$ are collinear.
It is not hard to see that is the external angle bisector. Let and be the intersections of with lines and respectively.
- Quadrilaterals $QHVB$ and $QHWC$ are both cyclic, and in particular tangent to line $BC$ .
- Now, a straightforward angle chase implies that $\angle HVF = \angle HBC$ and $\angle HWE = \angle HCB$ which is the key result in 2021 EGMO Problem 3.
- Returning to the previous spiral similarity, this means that lines $HV$ and $H_bD$ as well as lines $HW$ and $H_cD$ concur on $(ABC)$ .
An easy angle chase implies that points , and are collinear.
- Thus, the aforementioned spiral similarity maps $P$ to $(ABC) \cap AH$ and $N$ to $AN\cap QH$ .
- Hence, $FP \parallel BN$ and $EP \parallel CN$ .
- Also this spiral similarity maps $H$ to the intersection of $QP$ with $BC$ so $QH \cap (AEF)$ lies on the perpendicular from $A$ to $EF$ , a genarlized version of which is given in this problem from China.
Some harmonic bundle chasing using the above claim then implies that quadrilateral is harmonic.
- Noting that the internal $\angle BAC-$ bisector is the perpendicular bisector of segment $EF$ we have that $NE=NF$ . Thus, this claim implies that $N$ is the intersection of the tangents to $(AEF)$ at $E$ and $F$ , finishing off 2019 Polish MO Problem 1.
- This also implies that lines $QP$ , $AX$ and $BC$ concur.
Throwing in the second intersection of line , we can obtain several claims via pure angle chasing.
- Quadrilaterals $BFHT$ , $CEHT$ , $FETN$ , $Q_aHPT$ and (pentagon) $Q_aHRH_aG$ are all concyclic.
- Line $TH$ can also be observed to be the internal $\angle ETF-$ bisector.
The line passes through several key points. For starters, an easy angle chase shows that it passes through the trapezoid point.
- After a rt-bc inversion, one can show that lines $BF$ and $CE$ concur on this line as well which solves the following very enjoyable problem.
A related configuration can be seen in this problem, where an analogue of the second bullet point in Claim 4 above holds, for the circumcenter.