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 5 of 7: F is the circumcenter of AMN

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\angle MHN+\angle MFN=180^\circ \Longrightarrow M,F,N,H\ \text{concyclic}; \qquad FA=FM=FN

Detailed analysis

By steps 3–4, $\angle MHN=180^\circ-2\angle BAC$ and $\angle MFN=2\angle BAC$ sum to $180^\circ$ , so $M,F,N,H$ lie on one circle. Since $FM=FN$ (equal base angles in step 4), $\angle MFN=2\angle BAC$ , and $F$ lies on the same side of $MN$ as $A$ , the point $F$ is exactly the circumcenter of triangle $AMN$ ; hence $FA=FM=FN$ .