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 3 of 4: Claim:

LA=LQ

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In plain words

A rectangle formed by midpoints of $HQ$ and $AH$ forces the circumcenter $O$ to lie on line $LT$ , which is exactly the perpendicular bisector of $AQ$ .

LA=LQ

Detailed analysis

Let $N$ and $T$ be the midpoints of $HQ$ and $AH$ , and let $O$ be the circumcenter of $ABC$ . Since $MT$ is a diameter of the nine-point circle (as $T$ is the midpoint of $AH$ and $M$ is the midpoint of $BC$ , a classical nine-point circle diameter), the quadrilateral $LTNM$ is a rectangle, so line $LT$ passes through the center $O$ of the original circumcircle. Since $LT\perp AQ$ (as $LT\parallel$ to the earlier perpendicular direction) and $OA=OQ$ (both circumradii), $O$ lies on the perpendicular bisector of $AQ$ , and since $L,T,O$ are collinear with $LT\perp AQ$ , point $L$ also lies on that perpendicular bisector. Hence $LA=LQ$ .