Problem 4

Let

ABC

be an acute-angled triangle with orthocenter

H

, and let

W

be a point on the side

BC

, lying strictly between

B

and

C

. The points

M

and

N

are the feet of the altitudes from

B

and

C

, respectively. Denote by

\omega_1

the circumcircle of

BWN

, and let

X

be the point on

\omega_1

such that

WX

is a diameter of

\omega_1

. Analogously, denote by

\omega_2

the circumcircle of

CWM

, and let

Y

be the point such that

WY

is a diameter of

\omega_2

. Prove that

X, Y

and

H

are collinear.

Step 4 of 5: A power-of-a-point identity involving H

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AL\cdot AH = AB\cdot AN = AW\cdot AZ

Detailed analysis

The quadrilateral $BLHN$ is cyclic, since $\angle BLH=\angle BNH=90^\circ$ are two opposite right angles. Computing the power of $A$ with respect to $\omega_1$ and to the circle through $B,L,H,N$ gives $AL\cdot AH=AB\cdot AN=AW\cdot AZ$ , where the last equality is the power of $A$ with respect to $\omega_1$ , using the radical-axis relation already established.