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 3 of 7: Use the orthocenter and the tangency

Previous step Next step

\angle ABH = 90^\circ-\angle BAC=\angle ACH,\qquad \angle MHN=\angle MBH+\angle NCH=180^\circ-2\angle BAC

Detailed analysis

Since $H$ is the orthocenter of $ABC$ , the altitude from $B$ is perpendicular to $AC$ , giving $\angle ABH=90^\circ-\angle BAC$ , and symmetrically $\angle ACH=90^\circ-\angle BAC$ . Because the circumcircles of $BMH$ and $CNH$ are tangent at $H$ , the tangent-chord angles at $H$ combine with these values to give $\angle MHN=\angle MBH+\angle NCH=180^\circ-2\angle BAC$ .