Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Wednesday, September 26, 2018

### Geometry Problem 1389: Triangle, 40-120-20 degree, Angle, Congruence

Labels:
40-120-20,
angle,
congruence,
degree,
geometry problem,
triangle

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Find E on AC such that AE = BE. Construct equilateral Tr. BEF, F within Tr. BCE

ReplyDeleteSince BC = AD = CE, AE = BE = EF = CD

So Tr.s CEF & BCD are congruent SAS

Hence < FCE = < CBD = 10 and so

x = 30

Sumith Peiris

Moratuwa

Sri Lanka

Let point E between A and D such as AE=BE.So <DBC=10.

ReplyDeleteLuăm punctul E pe prelungirea lui BC astfel încât CE = CA; triunghiul AEC este isoscel (80-20-80).

ReplyDeleteDucem BF paralel cu EA și atunci se arată ușor că EB=BF=FA=CD(triunghiul ABF isoscel 40-100-40).

Construim un triunghi isoscel CDP cu unghiul DPC = 20.

Deoarece CD = BF, implică triunghiul CBF = CDP și de aici CP = CB. De asemeni unghiul BCP = 60, ceea ce face triunghiul BPC echilateral(60-60-60) și PB =PC=PD. În triunghiul BDP isoscel BP = DP și

unghiul BPD =40. De aici unghiul PDB = 70, adică unghiul BDC = 150 și deci BDA =30.

Florin Popa , Comanesti, Romania.

Considering usual triangle notations,

ReplyDeletelet BC=AD=a and DC=x

Mark a point E on AC such that CE=a => AE=x ------------(1)

Observe that BEC is an 80-80-20 isosceles and ABE is a 40-40-100 isosceles with AE=BE=x -----------(2)

Now extend the internal angle bisector of m(ABC) to a point P such that BP=a and this forms an equilateral triangle BPC

From this construction it can be observed that C is the circumcenter of EBP

=> m(BEP) = (360-m(BCP))/2 = 360-60/2 = 150 degrees----------(3)

Another observation is that EB=DC=x, m(EBP)=m(DCB)=20 & BP=BC=a => EBP is congruent to DCB (SAS)

=>m(CDB)=m(BEP)=150

=>m(BDA)=x=30 Q.E.D

Extend (AB to E so that triangle BCE be equilateral, take F on AC so that EF=EC, we get <EFC=<FCE=80, and <BEF=40, thus AF=EF and F=D, hence <AFB=<BEC/2=30.

ReplyDelete