Cho ABC. Về  phía  ngoài  của  tam  giác  ta  dựng  các  tam  giác  vuông  cân  ABE  và  ACF  đỉnh  A.Chứng  minh  rằng  trung  tuyến  AI  của  tam  giác  ABC  thì  vuông  góc  với  EF  và  AI=12EF
Gọi K là điểm đối xứng với A qua IQ là giao điểm của AIEF
Xét CIKBIA có 
IB=IC,IA=IK  ^AIB=^CIK
CIK=BIA(c.g.c)CK=AB,CK//AB
Do ^EAB=^FAC=90^EAF+^BAC=180
Do CK//AB^ACK+^BAC=180
Vậy ^EAF=^ACK
Kế hợp với CK=AE   ( cùng bằng AB ) , AC=AF
Ta suy ra AEF=CEK(c.g.c) 
AK=EFAI=12EF
^CAK=^AFE
Mà  ^CAK+^FAQ=90  do ^FAC=90
^AFE+^FAQ=90AQ vuông góc EF

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