Challenging Geometry Puzzle: Problem 1554. 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 BH to D and E such that < ADH = 80 and < AEH = 50.

ReplyDeleteExtend BC to F such that < AFH = 40

Now Triangle EAF is Right Angled with B as centre

Hence AD = AB = BF = BE = c

So CF = c - a

BD = 2 X (4 - a/2) = 8 - a and DE = a (since Triangles ADE & ABC are congruent ASA)

Hence c = BE = BD + DE = (8 - a) + a = 8

Sumith Peiris

Moratuwa

Sri Lanka

Slightly modified solution

ReplyDeleteDraw a Circle AEF with Centre B and Radius BA so that EHBMCF is a Diameter. Mark D on HE such that DH = BH = 4 - a/2 and AD = AB = c

Triangles ABD & ADE are both isosceles and Triangles ADH and ABC are congruent ASA

So DE = BC = a and BD = c - a

But BD = BH X 2 = (4 - a/2) X 2 = 8 - a

Hence BD = c - a = 8 - a and c = 8 = AB

Sumith Peiris

Moratuwa

Sri Lanka

Amazing..... Great Sri lankan son

ReplyDeleteLet the bisector of angle ABC intersect AC at D.

ReplyDeleteSince <DCB = <DBC= 50, DBC will be an isosceles triangle.

Draw DM which will be parallel to AH.

From similarity, CM/MH = CD/DA.

And from angle bisector theorem BC/BA=CD/DA

Hence, CM/MH = BC/BA

Since BC = 2 CM, BA = 2 MH = 2*4 = 8.