Question 62347
please help me slove this proof problem ---> "Prove that the exterior vertex angel is congruent to the central angle of a regular ploygon. [using congruence and or equal angle measure relationships, not by formula]
IN A REGULAR POLYGON ALL SIDES ARE EQUAL.ALL INTERNAL ANGLES ARE SAME.TAKE ANY 2 ADJACENT SIDES AB AND BC.LET O BE THE CENTRE OF THE POLYGON.JOIN OA,OB AND OC.
OA=OB=OC.......ANGLE
HENCE IN TRIANGLES OAB AND OBC,
OB = OB
OA = OC 
AB = BC
HENCE THE 2 TRIANGLES ARE CONGRUENT.
ANGLE OBA= ANGLE OBC = X SAY.................1
SO ANGLE ABC = ANGLE ABO + ANGLE OBC = 2X........2
SINCE OA=OB 
ANGLE OAB = ANGLE OBA = X .....................3
HENCE VERTEX ANGLE = ANGLE AOB = 180 - ANGLE OBA - ANGLE OAB = 180-2X......4
IF AB IS EXTENDED TO Z THEN 
ANGLE OBZ IS THE EXTERNAL ANGLE IN TRIANGLE OAB = SUM OF OPPOSITE INTERNAL ANGLES = ANGLE OAB + ANGLE OBA = X + 180 - 2X = 180 - X.
EXTERNAL ANGLE OF POLYGON = ANGLE CBZ = ANLE OBZ - ANGLE OBC 
= 180 - X - X = 180 - 2X = VERTEX ANGLE OF POLYGON.