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<CBQ = <PBA wanted, feet of altitudes and circumcircle related

Source: 2021 Cono Sur Shortlist G5 https://artofproblemsolving.com/community/c1088686_cono_sur_shortlist__geometry

October 31, 2022

geometryequal angles

Problem Statement

Let

\vartriangle ABC

be a triangle with circumcenter

O

, orthocenter

H

, and circumcircle

\omega

.

AA'

,

BB'

and

CC'

are altitudes of

\vartriangle ABC

with

A'

in

BC

,

B'

in

AC

and

C'

in

AB

.

P

is a point on the segment

AA'

. The perpenicular line to

B'C'

from

P

intersects

BC

at

K

.

AA'

intersects

\omega

at

M \ne A

. The lines

MK

and

AO

intersect at

Q

. Prove that

\angle CBQ = \angle PBA

.

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