Challenging Geometry Problem 1611. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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Thanks for the hint Antonio
ReplyDeleteThe perpendicular from C to AD extended meets at E. CE extended meets AB extended at F
< FBC = 73 = < ACF = < AFC
So FC = CB = CD = DF, hence Triangle CDF is equilateral
Hence < BCD = 60 - 34 = 26 and
x = 39 - 26 = 13
Sumith Peiris
Moratuwa
Sri Lanka