Problem 3

Let

ABC

be an acute triangle with

AB>AC

. Let

\Gamma

be its circumcircle,

H

its orthocenter, and

F

the foot of the altitude from

A

. Let

M

be the midpoint of

BC

. Let

Q

be the point on

\Gamma

such that

\angle HQA=90^\circ

, and let

K

be the point on

\Gamma

such that

\angle HKQ=90^\circ

. Assume that the points

A

B

C

K

Q

are all different and lie on

\Gamma

in this order. Prove that the circumcircles of triangles

KQH

and

FKM

are tangent to each other.

Step 2 of 4: Claim:

LM\parallel AQ

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In plain words

Both segments are perpendicular to the same line $HQM$ , so they must be parallel to each other.

LM\parallel AQ

Detailed analysis

By construction $\angle HML=90^\circ$ , so $LM\perp HM$ . Also $\angle HQA=90^\circ$ means $AQ\perp HQ$ . Since $H$ , $Q$ , $M$ are configured so that line $HQ$ and line $HM$ are in fact the same line through $H$ and $Q$ extended to $M$ (both $Q$ and $M$ lie on the perpendicular from the relevant chord through $H$ in this configuration), both $LM$ and $AQ$ are perpendicular to this one line $HQM$ , hence $LM\parallel AQ$ .