Challenging Geometry Puzzle: Problem 1577. 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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Extend AC upto E such that DC = CE = p say
ReplyDeleteLet BD extended meet AC at F
< CDE = < DEC = 40/2 =20 = < DBC
So BDCE is concyclic
Reference my Proof of Problem 1576,
AC = DE = BD = b
So < BED = 40 & < BEA = 60 = < FDC giving
< BFC = < BCF = 80
Hence Triangles BFC and DFE are both 80-20-80 Triangles with
BF = BC and
DE = FE = b
So CF = b - p and so
AF = p
So in Triangles ABF & BCE
AF = CE = p
BF = BC and
< BFA = < BCE = 100
So Triangles ABF and CBE are congruent SAS and
AB = AE
Since < AEC = 60 it follows that Triangle ABE is equilateral
Hence AB = AE = b + p = AC + CE and hence
AB = BD + CD
Sumith Peiris
Moratuwa
Sri Lanka
My solution, very slightly different from above, at https://stanfulger.blogspot.com/2024/09/gogeometry-1577.html
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