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 5 of 5: Conclusion: H lies on line XYZ

Previous step Next step

\triangle AHZ \sim \triangle AWL \implies \angle HZA = 90^\circ

Detailed analysis

If $H$ lies on line $AW$ , the identity $AL\cdot AH=AW\cdot AZ$ forces $H=Z$ directly. Otherwise, the equal ratios $\frac{AH}{AZ}=\frac{AW}{AL}$ together with the common angle at $A$ make triangles $AHZ$ and $AWL$ similar, so $\angle HZA=\angle WLA=90^\circ$ ; thus $H$ also lies on the line through $Z$ perpendicular to $AZ$ , which is exactly the line $XYZ$ , proving $X,Y,H$ are collinear.