See complete Problem 109

Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 26, 2008

### Geometry Problem 109

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## Monday, May 26, 2008

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Geometry Problem 109

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See complete Problem 109

Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

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Construct E on AC where AE=BE, then BCE is isosceles. Let AE=BE=CE=1 and AB=x, then by angle bisector theorem DE=1/(x+1), so 1/(x+1)+1=x, giving x=sqrt(2). Thus angle BAC is pi/4, therefore also angle ACB.

ReplyDeleteIf in the construction suggested by Anonymous AE = BE then Tr. AEB is isosceles and BD bisects angle ABE. That's easily understood but why should Tr. BCE be isosceles too?

ReplyDeleteE is centre of right Tr. ABC, that's why.

Deletehttp://geometri-problemleri.blogspot.com/2009/05/problem9-solution-on-10-may.html

ReplyDeleteThe trigonometric solution here is interestingly simple for tan(2α) = BC/AB = BC/CD = sin(3α)/cos(α) applying the Sine Rule in Tr. BCD. Thus sin(2α)/cos(2α) = sin(3α)/cos(α) leading to sin(3α)=sin(5α) which is possible only when α=22.5 deg. ruling out α=67.5 since 2α<90 deg. and thus x = 2α = 45 deg.

ReplyDeleteThe hint about requiring a construction was very helpful. I thought about bisecting angle A to get an isosceles triangle with angles 'alpha' but that seemed to make the figure very complex. I then thought of translating AB such that A maps to D and B to a point E, say, to make another isosceles triangle CDE. It then turns out that CDBE is cyclic and this leads to a proof...

ReplyDeleteLet O be the midpoint of AC the centre of circle ABC

ReplyDeleteOB = b/2. AD = b-c and DO = c-b/2

Since BD bisects < ABO

c/b/2 =(b-c)/(c-b/2) and so b^2 = 2c^2

Hence a= c and x = 45

Sumith Peiris

Moratuwa

Sri Lanka

By External Angle Theorem: m∠BDC=3α. Let E be a point on BC such that DE∥AB, we have m∠ABC=m∠DEC=90°, m∠A=m∠EDC=2α, m∠ABD=m∠BDE=α, they are pairs of corresponding angles.

ReplyDeleteLet F be a point on BC such that DF is the angle bisector ∠EDC, by Leg – Angle theorem: ∆BED≅∆FED, it follows that BD=DF.

Since AB=DC, by SAS: ∆ABD≅∆CDF, therefore: ∠A≅∠C,x=2α, we have x+2α=x+x=90°,x=45°.

<DBC=90-a; take E onto BC so that <BDE=2a. Clearly tr. BDE is isosceles and <CDE=<ABD=a, thus triangles BAD and DCE are congruent, s.a.s., wherefrom <DCE=<BAD, i.e. 2x=90, x=45.

ReplyDelete[For easy typing, I use "a" instead of alpha]

ReplyDelete<DBC=90-a & x=90-2a

Consider triangle BCD

sin(90-a)/CD=sinx/BD

CD/BD=sin(90-a)/sinx=cosa/cos2a---------(1)

Consider triangle ABD

sin2a/BD=sin(180-3a)/AB

AB/BD=sin3a/sin2a-------(2)

Since AB=CD, by equating (1) & (2)

cosa/cos2a=sin3a/sin2a

sin3acos2a=sin2acosa

sin5a+sina=sin3a+sina

sin5a=sin3a

5a=3a or 5a=180-3a

a=0 (rej) or 22.5

x=90-2a=45