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 4 of 4: Conclude tangency

Previous step Next step

In plain words

A chord parallel to the tangent line at the point diametrically balanced by equal radii is exactly the tangent-chord angle condition.

LM\parallel AQ,\ LA=LQ \Rightarrow ML \text{ tangent to } (AQL)

Detailed analysis

Since $LA=LQ$ , triangle $LAQ$ is isosceles, so the tangent to its circumcircle $(AQL)$ at $L$ makes with chord $LQ$ the same angle that $LA$ makes with $LQ$ on the other side (tangent-chord angle equals the inscribed angle $\angle LAQ$ in the alternate segment, which for the isosceles triangle equals $\angle LQA$ ). Combined with $LM\parallel AQ$ , this forces line $ML$ itself to be the tangent to $(AQL)$ at $L$ . Undoing the inversion $\iota$ , the tangency of line $ML$ to circle $(AQL)$ at $H$ 's inverse point $L$ translates back to the tangency of circle $(KQH)$ (inverse of line $ML$ through the center $H$ ) and circle $(FKM)$ (inverse of $(AQL)$ ) at their common point $K$ , which is exactly what was to be proved.