## Tuesday, February 24, 2009

### Problem 256: Equilateral Triangle, Circumcircle, Point, Vertices, Distances

Proposed Problem

See complete Problem 256 at:
gogeometry.com/problem/p256_equilateral_triangle_circumcircle_distance_vertex.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

1. Use Ptolemy's therem by which in this case we've:
AD*BC = CA*BD + AB*CD or d*BC = e*CA + f*AB which leads us to d = e + f since AB = BC = CA
Ajit: ajitathle@gmail.com

2. according to ptolemy's theorem, AB.CD + BD.AC = BC.AD. but triangle ABC is equilateral. so AB = BC = CA so we have AD = BD + CD that is d = e + f
Q. E. D.

3. Obviously ∠BDA = ∠ADC = 60°.

Rotate 60° anti-clockwise about D.
Let B→B'. Then ΔBDB' is equilateral, and DB' = e.

Rotate 60° anti-clockwise about B.
Then B'→D, A→C, so B'A = DC = f.

As a result, d = DA = DB' + B'A = e + f.

4. Problem 256 (solution 1)
Following on the DB to point B take the point E such that DE=AD (<EDA=<BDA=<BCA=60).Then triangleADE is equilateral .But triangleAEb=triangleADC
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5. Problem 256 (solution 2)
Following on the BD to point D take the point F such that DF=DC (<BDA=<CDA=<CDF=60).Then triangleCDF is equilateral .But triangleBCF=triangleADC
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6. Problem 256 (solution 3)
On the side AB ( not to extend ) get the point K such that BD=BK.Then triangle BKD is equilateral(<BDK=60=
<BKD ),so triangleABK=triangleCBD (AB=BC,BK=BD,<ABK=60-KBC=<CBD), so AK=DC. Therefore AD=AK+KD=DC+DB.
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7. Extend CD to F such that BF//AD.
Then Tr. BDF is equilateral and
Tr.s ABD ≡ CBF.

Hence AD = CF = CD + DF so d = e + f

Sumith Peiris
Moratuwa
Sri Lanka