Problem 2

Let

ABC

be an acute-angled triangle with

AB<AC

. Let

\Omega

be the circumcircle of

ABC

. Let

S

be the midpoint of the arc

CB

\Omega

containing

A

. The perpendicular from

A

BC

meets

BS

D

and meets

\Omega

again at

E\neq A

. The line through

D

parallel to

BC

meets line

BE

L

. Denote the circumcircle of triangle

BDL

\omega

. Let

\omega

meet

\Omega

again at

P\neq B

. Prove that the line tangent to

\omega

P

meets line

BS

on the internal angle bisector of

\angle BAC

Step 5 of 6: Combine the diameter and the parallel line

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In plain words

A diameter always sees any point on the circle at a right angle, so FP is perpendicular to AP; once OH is parallel to FP, it must also be perpendicular to AP, which places H on the perpendicular bisector of AP.

HP=HA

Detailed analysis

Because $AF$ is a diameter of $\Omega$ (Step 3), $\angle APF=90^\circ$ , i.e. $FP\perp AP$ . Since $OH\parallel PD=PF$ (Step 4), it follows $OH\perp AP$ . As $O$ is already equidistant from $A$ and $P$ (both on $\Omega$ ), the perpendicular to $AP$ through $O$ is exactly the perpendicular bisector of $AP$ , and $H$ lies on this line too, so $HP=HA$ .