See complete Problem 7 at:
www.gogeometry.com/problem/problem007.htm
Triangle, Cevian, Angles, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Sunday, May 18, 2008
Elearn Geometry Problem 7
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trazar la ceviana exterior BE (que corta a la prolongacion de AC en E)
ReplyDeleteSi el angulo EBC=x
entonces el angulo BEC=3x...(1)
Ademas los triangulos ABC y BED son congruentes
y de (1) se tiene que: el angulo ABC=3x
Finalmente en el triangulo ABC
3x+4x+3x=180º
10x=180º
x=18º
atte:tino
http://geometri-problemleri.blogspot.com/2009/10/problem-34-ve-cozumu.html
ReplyDeletehttp://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf
ReplyDeleteGiven:
ReplyDeleteABC=A+B+C
ABC=180 A=3x B=x+2x C=4x
Solution:
ABC=A+B+C
180=3x+x+2x+4x
180=10x
180/10=10x/10
18=x
answer:
A=3x B=x+2x C=4x
A=3(18) B=18+2(18) C=4(18)
A=54 B=18+36 C=72
B=54
180=54+54+72
180=180
To Jonathan (Problem 7): Why B=x+2x?
ReplyDeleteVideo-solution to the problem by me:
ReplyDeletehttp://www.youtube.com/watch?v=cMkCjlKlgRk
Greetings.
The solution is uploaded to the follwing link
ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJLWMyd3E5THpTTTJKZHdOaVU2UjRzZw
Triangles in the first step can not be congruent by AAS
Deletethere is no criteria of AAS for congruence
There is no rule of AAS congruence, side common between angles must be same...................
DeleteSea E el punto en la prolongación de A-C tal que CE = AD
ReplyDeleteSea F tal que F-E es paralela a D-B y además EF=DB
Con esto se arma un rombo DBFE
La diagonal B-E corta a C-F en un punto G
Esta diagonal también es bisectriz por lo que ang(GEC)=3α, lo que hace que
△CGE y △BGF sean isósceles.
Como BFEC es un paralelogramo recto se deduce que:
ang(ECB) = ang(FEC)
Pero esto implica:
6α=180°-4α
α=18° Q.E.D.
Construct isoceles Tr. ABE where E is on AC. Then Tr. BDE is isoceles and so DE = BD= AC
ReplyDeleteHence Tr.s ABC & BED are congruent SAS from which it follows that Tr. BDC is isoceles with internal angles adding upto 10x and so x=18
Sumith Peiris
Moratuwa
Sri Lanka
Construct point E such that BA=BE (hence triangle ABE is formed and it is an isosceles triangle, where <BAE=<BEA=3x). Additionally, E lies on AC produced.
ReplyDeleteConsider triangle ABC and EBD.
<ABD=180-9x
<ABE=180-6x
Hence <EBC=(180-6x)-2x-(180-9x)
<EBC=x
Hence <DBE=3x
Since <DBE=<DEB=3x, DB=DE
Since AC=DB (given), AC=DE
1.BA=BE (from above)
2.<BAC=<BED=3x (proven)
3.AC=ED (proven)
Hence triangle ABC is congruent to EBD (SAS)
Hence, BD=BC
<BDC=4x
In triangle BCD,
2x+4x+4x=180
x=18
By the law of sines,
ReplyDeletesin(∠BDC):sin(∠C) = BC:BD = BC:AC = sin(∠A):sin(∠ABC).
That is,
sin(180°-6x):sin(4x) = sin(3x):sin(180°-7x),
thus,
sin(6x):sin(4x) = sin(3x):sin(7x).
This implies
sin(7x)sin(6x) = sin(4x)sin(3x)
By the product-to-sum identities,
-(1/2)(cos(13x) - cos(x)) = -(1/2)(cos(7x) - cos(x)),
which can be reduced to
cos(13x) - cos(7x) = 0 (since cos(x) = 0 is not valid).
Then by the sum-to-product identity,
-2sin(10x)sin(3x) = 0
which suggests
sin(10x) = 0 or sin(3x) = 0.
This means x is either a multiple of 18° or a multiple of 60°.
Since ∠A = 3x, ∠C = 4x, and ∠ABC = 180°-7x must all be in (0°, 180°), we must have x = 18°.
See video
ReplyDeleteConstruct H projection of B on AC
ReplyDeleteConstruct E on AC such as DE=AC=BD
Then C and D are symmetrical with respect to H, thus <BDC=<BCD=4x, finally, in isoceles triangle ∆DBC:
180°=4x+4x+2x=10x and x=18°
Grab a pen and a piece of paper, and illustrate along.
ReplyDeleteFirstly, add point E to the figure and connect it to points B and C such that points A, C and E are collinear and angle BEC = 3x.
We can calculate that:
> Angle BCE = 180° - 4x
> Angle CBE = 180° - (180° - 4x) - 3x = x
> Angle DBE = 2x + x = 3x = angle BEC
> Triangle BDE is isosceles because angle BEC = angle DBE
> Line DE = line BD since triangle BDE is isosceles.
> Line DE = line BD = line AC
> Triangle ABE is isosceles since angles BAD and BEC are equal.
> Line AB = line BE since triangle ABE is isosceles.
Now we solve the problem:
> Line AE - line AC = line AE - line DE
> Line CE = line AD
> Triangles ABD and BCE are congruent by SAS:
· Line AB = line BE
· Angle BAD = angle BEC
· Line AD = line CE
> Angle ABD = 180° - 2x - 3x - 4x = 180° - 9x
> Since triangles ABD and BCE are congruent, angles ABD and CBE are equal.
> 180° - 9x = x
> 10x = 180°
Therefore x = 180°/10 = 18°
Construct point E such that AB=BE
ReplyDeleteSo,<BAD=<BEC=3x
Then,<CBE=x and <DBE=2x+x=3x
In ΔΒCE,
<DBE=<DEB=3x
So,BD=DE
Since AC=BD
We can say that AC=BD=DE
Now,
AE-DE=AE-AC
AD=CE
In ΔABD and ΔBCE,
AB=BE
AD=CE
<BAD=<BEC
We get these two triangles congruent.
So,<CBE=x=<ABD
Hence,<ABC=x+2x=3x
In triangle ABC,
3x+3x+4x=180
x=18 and we are done
Partho Saha
Bangladesh
Email: parthosaha.abir@gmail.com