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 4 of 7: T is the arc midpoint of EF on the circumcircle of BEF

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A':=CE\cap DF,\qquad T:=\text{circumcentre of }\triangle A'EF

Detailed analysis

The parallelogram $ACA'D$ gives $A'C\parallel AD$ and $A'D\parallel AC$ , while $F,A',D$ are collinear and $E,F,A$ are collinear. Therefore the three angles of $\triangle A'EF$ are explicitly $\angle FEA'=\angle AEC=\angle ABC=\angle CAB=90^\circ-\alpha$ , $\angle A'FE=\angle DFA=\angle DBA=\angle BAD=90^\circ-\beta$ , and $\angle EA'F=180^\circ-(90^\circ-\alpha)-(90^\circ-\beta)=\alpha+\beta$ . If $T$ is its circumcentre, then $\angle EA'T=90^\circ-\angle A'FE=\beta=\angle A'CD=\angle CA'B$ ; hence $A',B,T$ are collinear. Moreover, $\angle ETF=2\angle EA'F=2(\alpha+\beta)$ , whereas $\angle EBF=\angle EBA+\angle ABF=\angle ECA+\angle ADF=\angle A'CA+\angle ADA'=2(\alpha+\beta)$ . Thus $B,E,F,T$ are concyclic. Since $T$ is the circumcentre of $\triangle A'EF$ , $TE=TF$ , so on this circle $T$ is the midpoint of the arc $EF$ not containing $B$ . The tangent at an arc midpoint is perpendicular to the radius $OT$ , while the chord $EF$ is also perpendicular to $OT$ ; therefore that tangent is parallel to $EF=AP$ .