2012 ELMO #1, https://artofproblemsolving.com/community/c6t48f6h486930_two_circles_concur_on_a_line

 Let DEF be the orthic triangle of triangle ABC and L,M the centers of the circles passing through D and being tangent at F to the circle (AEF), respectively passing through D and being tangent at E to the circle (AEF). Prove that the circles L, M have BC as radical axis.

Proof:
We need to show that LM_|_BC. Let H be the orthocenter of tr. ABC abd K midpoint of AH. Easy angle calculation show that triangles DFL and DEM are similar, with base angles equal to 90-|B-C|, and one of them only has common points with interior of tr. DEF, consequently, constructing the isosceles triangle FEN, with N and D on either side of EF and similar to DFL and DEM, we know that LFNM is a parallelogram, but easy angle calculation shows FN_|_BC, thus we are done.

Floor van Lamoen, https://www.facebook.com/photo?fbid=1698024093723819&set=gm.4117748638338811

 Given arbelos A-C-B, D, E centers of (AC), (CB) respectively, while F, G are highest points of (AC), (CB) respectively. Prove that <AGD=<BFE.

Proof

Let V=AF x BG ( by homothecy it is highest point of (AB). Notice easy that FCGV is a rectangle, also if O is midpoint of AB, then DE=OA. Another interesting fact:
GV/FD=FC/FD=AV/AO=AV/DE=sqrt (2), meaning that triangles AGV and EFD are similar, so <VAG=<DEF, i.e. if T=AG x EF, <ATF=45 ( 1 )and T lies onto (AC). In a similar way, if U=BF x DG, U lies onto (CB) and <BUG=45 ( 2 ). From (1) and (2) we get FTUG cyclic, thus we are done.

Pablo Miguel Sancerni, reposted, Training Geometry Olympiad https://www.facebook.com/hacermatemagicas/photos/a.427762750694883/1026216000849552

ABCD is a square, ABE and CMN are equilateral triangles inside the square, E onto CM. AE with MN intersect at Q, BE with NC intersect at P. Prove that PQ_|_CN.

Proof:

<DCM=<BCN=15 (to get <MCN=60 and CM=CN), thus <AMN=<ANM=45. Next ANEM and CENB are cyclic, so AE is angle bisector of <MEN=90, <NEB=<NCB=15, <CNE=<CBE=30 and easy angles calculations give <MQE=75, <CPE=45.
As <PEQ=(<NQP+<NPQ)/2=60, E , onto the bisector of <MNC is the N-excenter of tr. NPQ, hence <EPQ=<EPC=45, consequently <CPQ=90, done.

Seiichi Kirikami, Mathematics today, https://www.facebook.com/photo?fbid=3118741878347510&set=gm.549309656439485

Let ABC be a triangle, I - its incenter, Ha, Hb, Hc the orthocenters of triangles BIC, CIA, AIB. Prove that AHb.BHc.CHa=AHc.BHa.CHb.

Proof:

AHb_|_CI_|_BHa, so <CAHb=<CBHa=90-<C/2 ( 1 ), similarly <ABHc=<ACHb ( 2 ) and
<BAHc=<BCHa ( 3 ). Applying sin theorem in triangle BCHa we get BHa/CHa=sin(<BCHa)/sin(<CBHa) ( 4 ), from triangle CAHb:
CHb/AHb=sin(<CAHb)/sin(<ACHb) ( 5 ) and from triangle ABHc: 
AHc/BHc=sin(<ABHc)/sin(<BAHc) ( 6 ).
Multiplying side by side relations (4)-(6) and taking into account relations (1)-(3) we get the desired relation.

Stanley Rabinowitz, Romantics of Geometry, https://www.facebook.com/photo?fbid=10223583965739224&set=gm.4106274612819547

 Given a segment AC and a random point B on it, construct the semicircles of diameters AB and AC and let F be midpoint of AB. A circle of center E is internally tangent to the semicircle (AC), externally tangent to semicircle (AB) and tangent to the tangent at B to semicircle (AB). Let the circle E touch the semicircle AB at P, and PF cut the tangent at B to semicircle (AB) at G. Prove that PG=BC.

Proof 

Note that the circle E is needed to get an idea of the position of P only: inversion of pole C and power CB.CA sends the semicircle (AC) to the line GB, while the semicircle (AB) and circle E remain unchanged, thus PC is tangent at P to semicircle (AB). As FP=FB ( 1 ) we infer that triangles FBG and FPC are congruent, FG=FC ( 2 ) and with (1) we get the desired PG=BC.

aops, 2019 Kharkiv Lyceum, https://artofproblemsolving.com/community/c4t48f4h2599212_bk2dc_if_ltabcltkadltakd_ltc90o_2019_kharkiv_lyceum_no_27__86

Let ABC be a C-right-angled triangle and D a point onto the side AC, K a point onto the segment BD so that <DAK=<AKD=<ABC. Prove that BK=2 CD.

Proof:

Clearly AC>BC and BC is tangent to the circle (ABK), this coming from <AKD=<ABC; let AC intersect this circle second time at E, further to C than A; from <DAK=<ABC it follows that
arc AE=arc KB, i.e. AE=KB and EBKA is an isosceles trapezoid, hence DE=DB. Take O the center of this circle and F midpoint of AE; DEOB is a rhombus ( a kite with OB||DE ), consequently OE||=DB, triangles EOF and DBC are congruent, EF=CD, thus BK=AE=2EF=2CD, done

 

Peru Geometrico, Stanley Rabinowitz https://www.facebook.com/photo?fbid=10222267784796604&set=gm.3897388763707321

 ABCD is a square, E a point onto AD, F onto BC so that BE||DF. CE x AB=G, GF x AD = H. Prove that <BEC=<FHE.

Proof:
Let AB=a. We have AE=CF ( 1 ), DE=BF ( 2 ). Take K = CH x BE, L = AF x BE.
From AD||BC: HE/AE=CF/BC; with (1), HE/AE=AE/a, giving a.HE=AE^2 ( 3 ). Divide both sides of (3) by a^2: HE/a=AE^2/a^2=AE^2/AB^2 ( 4 ), but HE/a=HE/BC=KE/BK ( 5 ), implying
KE/BK=(AE/AB)^2, thus AK_|_BE ( 6 ).
From similar triangles AED and ALE: AE/AD=AL/AF ( 7 ), while from triangles ALE and FLB similar we get AL/AF=EL/BE ( 8 ). But again HE/AH=CF/BF=AE/BF=EL/BL, hence LH||AB ( 9 ). From (6) and (9) ALKH is cyclic, so <ALE=<KHE, but <ALE=<CEB, done.