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

Labels:
angle,
cevian,
congruence,
triangle

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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