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H is the foot of the altitude on AB, such that AH=3HB

Source: Itamo 2014 - p2

November 19, 2014

symmetrygeometrycircumcirclegeometry unsolved

Problem Statement

Let

ABC

be a triangle. Let

H

be the foot of the altitude from

C

on

P

. Suppose that

AH = 3HB

. Suppose in addition we are given that
(a)

M

is the midpoint of

AB

; (b)

N

is the midpoint of

AC

; (c)

P

is a point on the opposite side of

B

with respect to the line

AC

such that

NP = NC

and

PC = CB

.
Prove that

\angle APM = \angle PBA

.

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