Problem 1

Let

H

be the orthocenter of triangle

ABC

. Let

M

and

N

be the midpoints of sides

AB

and

AC

, respectively. Assume that

H

lies inside quadrilateral

BMNC

and that the circumcircles of triangles

BMH

and

CNH

are tangent to each other. The line through

H

parallel to

BC

intersects the circumcircles of triangles

BMH

and

CNH

at points

K

and

L

, respectively. Let

F

be the intersection point of

MK

and

NL

, and let

J

be the incenter of triangle

MHN

. Prove that

FJ = FA

Step 6 of 7: Apply the lemma to triangle MHN

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H, J, F\ \text{are collinear}, \qquad FJ=FM=FN

Detailed analysis

Since $MFNH$ is cyclic with $FM=FN$ and $F$ on the arc not containing $H$ , $F$ is the midpoint of arc $MN$ not containing $H$ , so line $HF$ bisects $\angle MHN$ ; as $J$ is the incenter of triangle $MHN$ , it lies on this bisector, so $H,J,F$ are collinear. Applying the lemma of step 1 to triangle $MHN$ with $X=H$ , $D=F$ , $I=J$ gives $FJ=FM=FN$ .