Challenging Geometry Puzzle: Problem 1576. 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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Trigonometry Solution

ReplyDeleteLet BD = CD = p and let AD = d

From Triangle ABD,

p/sin30 = d/sin70 ................(1)

From Triangle BCD,

a = 2p cos20 = 2p sin70 ............(2)

(1) X (2) gives,

ap/sin30 = 2pd

So a = 2dsin30 = d

Sumith Peiris

Moratuwa

Sri Lanka

Pure Geometry Solution

ReplyDeleteExtend BD to E such that BD = DE, let BD = DC = DE = p

Let DF be perpendicular to AB, F on AB extended

Let AD = d

Note that ADF is a 30-60-90 Triangle and so DF = d/2

Now < BCE = 90 and Triangles BFD and BCE are similar

Hence DF / BD = BC / BE which gives

(d/2)/p = a/ (2p)

Therefore d = a, that is AD = BC

Sumith Peiris

Moratuwa

Sri Lanka