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 1 of 7: An incenter–arc-midpoint lemma

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DI=DY=DZ

Detailed analysis

For a triangle $XYZ$ with incenter $I$ , let $D$ be the second point where the internal bisector of $\angle YXZ$ meets the circumcircle of $XYZ$ , i.e. the midpoint of arc $YZ$ not containing $X$ . Since $\angle DIY$ is an exterior angle of triangle $IXY$ at $I$ , $\angle DIY=\angle IXY+\angle IYX=\tfrac12\angle X+\tfrac12\angle Y$ . Also $\angle DYZ=\angle DXZ=\tfrac12\angle X$ (same arc $DZ$ ) and $\angle ZYI=\tfrac12\angle Y$ , so $\angle DYI=\angle DYZ+\angle ZYI=\tfrac12\angle X+\tfrac12\angle Y=\angle DIY$ . Hence triangle $DIY$ is isosceles with $DI=DY$ , and the symmetric argument gives $DI=DZ$ . We apply this later to triangle $MHN$ with $D=F$ and $I=J$ .