Problem 2

Let

\Omega

and

\Gamma

be circles with centres

M

and

N

, respectively, such that the radius of

\Omega

is less than the radius of

\Gamma

. Suppose circles

\Omega

and

\Gamma

intersect at two distinct points

A

and

B

. Line

MN

intersects

\Omega

C

and

\Gamma

D

, such that points

C

M

N

and

D

lie on the line in that order. Let

P

be the circumcentre of triangle

ACD

. Line

AP

intersects

\Omega

again at

E\ne A

. Line

AP

intersects

\Gamma

again at

F\ne A

. Let

H

be the orthocentre of triangle

PMN

. Prove that the line through

H

parallel to

AP

is tangent to the circumcircle of triangle

BEF

Step 6 of 7: The orthocenter H inherits the same two directions

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MH\parallel AD,\qquad NH\parallel AC

Detailed analysis

Both $P$ (since $PA=PD$ ) and $N$ (the centre of $\Gamma\ni A,D$ ) lie on the perpendicular bisector of chord $AD$ , so line $PN$ is exactly that perpendicular bisector and $PN\perp AD$ . The altitude $MH$ of $\triangle PMN$ from $M$ is by definition perpendicular to $PN$ , hence $MH\parallel AD$ . Symmetrically, $P$ (since $PA=PC$ ) and $M$ (the centre of $\Omega\ni A,C$ ) both lie on the perpendicular bisector of chord $AC$ , so $PM\perp AC$ , and the altitude $NH\perp PM$ gives $NH\parallel AC$ .