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< NA'T = < ADT wanted, starting with a right triangle, symmetric, projections

Source: JBMO Shortlist 2018 G2

July 22, 2019

geometryequal anglessymmetryright triangle

Problem Statement

Let

ABC

be a right angled triangle with

\angle A = 90^o

and

AD

its altitude. We draw parallel lines from

D

to the vertical sides of the triangle and we call

E, Z

their points of intersection with

AB

and

AC

respectively. The parallel line from

C

to

EZ

intersects the line

AB

at the point

N

. Let

A'

be the symmetric of

T

with respect to the line

IK

and

I, K

the projections of

A'

onto

AB

and

AC

respectively. If

T

is the point of intersection of the lines

IK

and

DE

, prove that

\angle NA'T = \angle ADT

.

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