Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 695.
Wednesday, November 30, 2011
Problem 695: Triangle, Angles, Auxiliary Construction, Congruence, Mind Map, Polya
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ReplyDeleteextend BA to K
ReplyDeleteBC bisector
BD=BK
Join C to K
CK=CD
see the KCDA cydric quatrilateral
x=25
Problem 695 again:
ReplyDeleteLet the circle through B, C, D cut AC at E.
∠EBD = ∠ECD = 27° and so ∠EBC = 90°.
Let O be the midpoint of EC.
OB = OE = OC and O is the centre of circle BCD
∠AOD = ∠EOD = 2∠ECD = 54° = ∠ABD
So A,B,O,D are concyclic.
Hence
x = ∠DAO = ∠DBO
= 63° - ∠OBC
= 63° - ∠OCB
= 63° - 38°
= 25°