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.

Jean-Louis - Square, https://artofproblemsolving.com/community/c6t48f6h2598255_a_bissector_and_a_square

 Let I be incenter of triangle ABD os the square ABCD, E the intersection of DI with AB and F the contact of the incircle with AD. Prove that FI is the angle bisector of <CFI.

Proof:
Easy angle calculation gives that triangles CDI and AEI are isosceles, so EA/CD=AI/CI=AF/FD, i.e. AE/CD=AF/DF, relation proving that triangles AEF and DCF are similar, done.

My solution to https://artofproblemsolving.com/community/c4t48f4h2595840_equal_angles_in_equilateral_grid__201617_savin_competition_79_p18

 In the attached grid made by unit regular triangles there are points A(0,1), B(4,0), C(2,1), D(3,5). Prove that <ABC=<CAD.

My proof:
Take the equilateral triangle XYZ, X(0,0), Y(5,0), Z(0,5) and the point E(1,4) onto YZ. Clearly triangle ABE is equilateral, since triangles XBA and ZAE are congruent (s.a.s.). Take now F(4,4) and notice trat triangles BFD and AZE are congruent, thus BD||=AE, ABDE is a rhombus with a 60 degs angle, therefore <BAD=30 degs=<CBX. With <ABX=<BAC we get the required <ABC=<CAD.

https://artofproblemsolving.com/community/c4t48f4h2595482_right_angle_equal_ratios_equilateral_201617_savin_competition_79_p9

 

https://www.facebook.com/photo?fbid=1665737510303997&set=gm.492856491974669

 

Stanley Rabinowitz, Peru Geometrico, https://www.facebook.com/photo/?fbid=10222304299109439&set=gm.3911649568947907

 

Test Serbia, 2021, solution with my son, dr. A.M. Fulger