Felipe Augusto (footballer, born 1993)

From Wikipedia, the free encyclopedia
(Redirected from Felipe Abreu)
Felipe Augusto
Personal information
Full name Felipe Augusto de Abreu
Date of birth (1993-08-31) 31 August 1993 (age 30)
Place of birth Rio de Janeiro, Brazil
Height 1.70 m (5 ft 7 in)
Position(s) Forward
Team information
Current team
Rio Tinto
Number 17
Youth career
2011 Serrano
2012 Braga
Senior career*
Years Team Apps (Gls)
2012 Vidago 10 (0)
2013 Tourizense 10 (0)
2013–2018 Vizela 123 (10)
2018–2019 Trofense 28 (5)
2019–2020 Varzim 1 (0)
2020Trofense (loan) 6 (0)
2021–2022 Ribeirão 13 (1)
2022 Serzedelo 8 (0)
2022– Rio Tinto 10 (0)
*Club domestic league appearances and goals, correct as of 18:22, 10 June 2023 (UTC)

Felipe Augusto de Abreu, known as Felipe Augusto or Felipe Abreu (born 31 August 1993) is a Brazilian football player who plays for Rio Tinto.

Club career[edit]

He made his professional debut in the Segunda Liga for Vizela on 6 August 2016 in a game against Académico de Viseu.[1]

References[edit]

  1. ^ "Game Report by Soccerway". Soccerway. 6 August 2016.

External links[edit]