MathLabs

Problem 1

Let HH be the orthocenter of triangle ABCABC. Let MM and NN be the midpoints of sides ABAB and ACAC, respectively. Assume that HH lies inside quadrilateral BMNCBMNC and that the circumcircles of triangles BMHBMH and CNHCNH are tangent to each other. The line through HH parallel to BCBC intersects the circumcircles of triangles BMHBMH and CNHCNH at points KK and LL, respectively. Let FF be the intersection point of MKMK and NLNL, and let JJ be the incenter of triangle MHNMHN. Prove that FJ=FAFJ = FA.
Step 4 of 7: Transfer the angle through the parallel chord
Previous stepNext step
∠MBH=∠MKH=∠NCH=∠NLH=90∘−∠BAC,∠FMN=∠FNM=90∘−∠BAC,∠MFN=2∠BAC\angle MBH=\angle MKH=\angle NCH=\angle NLH=90^\circ-\angle BAC,\qquad \angle FMN=\angle FNM=90^\circ-\angle BAC,\qquad \angle MFN=2\angle BAC
Detailed analysis

Since M,B,H,KM,B,H,K lie on one circle, ∠MKH=∠MBH\angle MKH=\angle MBH, and since N,C,H,LN,C,H,L lie on one circle, ∠NLH=∠NCH\angle NLH=\angle NCH; combined with step 3 these four angles equal 90∘−∠BAC90^\circ-\angle BAC. The midline MNMN and the line KLKL through HH are both parallel to BCBC, so MN∥KLMN\parallel KL, hence the angles that MKMK and NLNL make with MNMN, namely ∠FMN\angle FMN and ∠FNM\angle FNM, equal 90∘−∠BAC90^\circ-\angle BAC too, giving ∠MFN=180∘−2(90∘−∠BAC)=2∠BAC\angle MFN=180^\circ-2(90^\circ-\angle BAC)=2\angle BAC.