T:
Problem #3
A:
WIlliam Meyer
G:
triangle ABC with angle bisectors segment AZ, segment BY, segment CX
triangle DEF with angle bisectors segment DJ, segment EI, segment FH
P:
angle bisectors in similar triangles same ratio as corresponding sides
Pr:
triangle ABC with altitudes segment AZ, segment BY, segment CX => G
triangle DEF with altitudes segment DJ, segment EI, segment FH => G
( m segment AB / m segment DE ) = ( m segment BC / m segment EF ) = ( m segment AC / m segment DF )
angle BAC congruent angle EDF => def similar triangles
angle ABC congruent angle DEF => def similar triangles
m angle ABY * 2 = m angle ABC => def angle bisector
m angle DEI * 2 = m angle DEF => def angle bisector
m angle DEI * 2 = m angle ABC => sub #-1, #-3
m angle DEI * 2 = m angle ABY * 2 => sub #-1, #-3
angle DEI congruent angle ABY => division & doca
triangle AYB similar triangle DIE => AA #-1, #-7
( m segment AB / m segment DE ) = ( m segment BY / m segment EI ) => similar triangles
( m segment BY / m segment EF ) = ( m segment BC / m segment EF ) = ( m segment AC / m segment DF ) => sub #-1, #-5
( m segment BC / m segment EF ) = (    m segment CX / m segment FH ) => similar reasoning #-2
( m segment AB / m segment DE ) = ( m segment CX / m segment FH ) = ( m segment AC / m segment DF ) => similar reasoning #-2
( m segment AC / m segment DF ) = ( m segment AZ / m segment DJ ) => similar reasoning #-4
( m segment AB / m segment DE ) = ( m segment BC / m segment EF ) = ( m segment AZ / m segment DJ ) => similar reasoning #-4