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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