Solution:Let, O be the orthocentre, circumcentre, incentre, and centroid ?ABC.So, AD is perpendicular on BC from vertex A? ? ADB = ? ADC = 90°Again, AD is the perpendicular bisector of the side BC? ?ADB = ?ADC = 90° and BD = CDAgain, AD is the bisector of the angle ? BAC? ?BAD = ?CADAnd, AD is also the median from vertex A to the side BC? BD = CD? From ?ABD and ?ACD?ADB = ?ADC = 90°?BAD = ?CADAnd AD is the common sideSo, ?ABD = ?ACD? AB = AC ¦.(1)Similarly, From ?ABE and ?CBE?ABE = ?CBE [ ? BE is the bisector of ? ABC ]?AEB = ?CEB = 90° [ ? BE is perpendicular on AC ]And BE is the common sideSo, ?ABE = ?CBE? AB = BC ¦.(2)? from (1) and (2)AB= BC = AC? ?ABC is equilateral