See complete Problem 6 at:

www.gogeometry.com/problem/problem006.htm

Triangle, cevian, equal segments, and angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Sunday, May 18, 2008

### Elearn Geometry Problem 6

Labels:
angle,
cevian,
congruence,
triangle

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CD/BD = sin(6x)/sin(3x)= 2cos(3x)

ReplyDeleteAB/BD = sin(3x)/sin(2x)

Thus, 2cos(3x) = sin(3x)/sin(2x)

or sin(2x) = (1/2)* sin(3x)/cos(3x)

Now if x=15 then sin(2x)=1/2 & sin(3x)=cos(3x) which satisfies the equation.

So x = 15 deg.

Ajit: ajitathle@gmail.com

http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

ReplyDeleteLet E be a point with triangles DBA and EDC are congruence(AB=CD,AD=CE,ang(BAD)=x=ang(DCE), ang(ACE)=4x). Then DB=DE and ang(EDC)=2x and hence

ReplyDeleteang(EBD)=x=ang(BED).

Furthermore, since BA=CD, BE=EC=AD=CA and ang(EBA)=3x=ang(ACD), triangles BAE and CDA are congruence. we see that AE=AD and so triangle AEC is an equilateral. we get 4x=60 degree, that is, x=15 degree.

Comment on "Joe": Trial-error is not accepted in the Netherlands

ReplyDeleteComment on "bae deaok rak":1)Where did you put point E? 2)AD=CA is not true

Given:

ReplyDeleteABC=180

A=2x

B=x+90(90 because that is right angle)

C=3x

Solution:

ABC=A+B+c

180=2x+x+90+3x

180=6x+90

-6x=90-180

-6x/-6=-90/-6

x=15

answer:

x=15+90 =2x =3x

=2(15) =3(15)

=30 =45

180=15+90+30+45

180=180

To Jonathan (Problem 6): In your statement B=x+90.... 90 is not a data.

ReplyDeleteif you notice the DBC this is 90%. and why B=x+90 because that is seperated. i'm right??

ReplyDeleteVideo solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 6. Thanks Eder and Cristian.

ReplyDeleteThe solution is uploaded to the following link:

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJWEZGVVpLQ2pRRUdza0E4NUoxODZUZw

To Rengaraj PradheepKannah

ReplyDeleteGreetings.

In your solution of problem 6, there’s something I didn’t understand.

The first circle is defined as having center at the midpoint G of AB (with diameter AB) and the second circle has center at the midpoint E of CD (with diameter CD).

I can’t see why the first circle meets CD in its midpoint E. If we don’t assure that E lies on the first circle, it’s not clear that AG = GE.

To Nilton Lapa

ReplyDeleteGreetings to you too.

Did you notice DBC is an Isosceles triangle.

Since E is the Mid point of DC BEC is 90 Deg.

As AB is the diameter of the first circle according to the angle on semi circle is 90 Deg

The first circle should go thriugh E.

Thanks.

PRadheepKannah

..............

Another way to clear up this confusion about the first circle passing through E:

Delete∠EDB is exterior of △ADB, so ∠EDB = ∠ABD + ∠DAB = x + 2x = 3x.

Therefore △DBC is isosceles (same base angles, ∠EDB = 3x = ∠BCE), that is, BD = BC.

Drop a perpendicular from B to DC (that is, BE⟂DC). As it is the altitude in an isosceles △DBC, it divides its base into equal pieces, so ED = EC and E is a midpoint of DC.

Now the part regarding the circle:

Note that △ABE is a right triangle, since BE⟂DC (and also BE⟂AC, because A,D,C are collinear).

Since we've chosen G to be the midpoint of its hypotenuse AB (AG = GB), the segment GE is a median in a right triangle △ABE, so it must also be equal to these parts: GA = GB = GE (three radii of the first circle).

Therefore points A, B, E lie on the same circle centered at G.

Rengaraj's response

assumesthat E lies on the circle when it says that it is a right triangle in a circle.To Rengaraj PradheepKannah

ReplyDeleteThanks. The explanation is absolutly clear. My doubt in problem 6 was enlighted.

Nilton Lapa

E es el punto sobre D-C tal que AD=DE

ReplyDeleteF es tal que B-F paralelo a A-C y BF=AD

D-B es paralela a E-F

G es intersección entre E-F y B-C formando EBFC cíclico

ang(DBC) = 180°-6α

△FCD es isósceles

(1) ang(FCD) =ang(DFC)=90°-α

△BGE es congruente a △FGC por la simetría

general en los triángulos isósceles △BFE y △EGC

y el paralelismo de BF con EC.

ang(CEB) = 90°-α

ang(EBF)=ang(BFC)=[360-2(90°-α)]/2=90°+α

ang(FHB)=180°-(90°+α)-2α=90°-3α

ang(EHF)=3α

Entonces △ABD es semejante a △EFH, por lo tanto

(2) ang(GEB)=ang(FCG)=2α

Comparando resutado (1) con (2)

ang(FCE)=90°-α=ang(FCG)+ang(BCA)=5α

6α=90°; α=15° Q.E.D.

Let E be a point on the extension AB such that AE=AC.

Bisect ∠A and let a point F on the angle bisector of ∠A such that m∠AEF=m∠ACF=x.

Then ∆AFE≅∆AFC by ASA and at the same time they are both isosceles.

Let a point H on BC such that the angle bisector of ∠A intersect BC at H.

Then m∠CFH=m∠EFH=m∠FEH=m∠FCG=2x, also m∠EBC=5x and m∠EGC=6x. An exterior angle is equal to the sum of its remote interior angles.

Again by ASA, ∆EHF≅∆CHF and they are also isosceles triangles, then EH=FH=CH.

By perpendicular bisector concurrence theorem, CG⊥EF that makes ∠EGC a right angle.

Therefore: 6x=90°,x=15°

What is G

ReplyDeleteCouldn't find the G spot, eh? ;)

DeleteJudging by the later statement about "CG⊥EF", I would say it must be the intersection point of EF and CB (just draw it in proportion and knowing the end result, and you'll see).

Still I don't quite follow the end of the proof, about using the "perpendicular bisector concurrence theorem", since nowhere has been proved that CH and EH are perpendicular bisectors of CF and EF (even if I can see it in the picture). The train of thoughts might be correct, but the proof is still incomplete.

See my proof, maybe similar to the video-one mentioned by the owner of the site on Sept, 4, 2011, but I am not able to see that.

ReplyDeleteTo Stan Fulger

DeleteSpanish solution video at http://www.gogeometry.com/geometria/p006_solucion_problema_video_triangulo_angulo_ceviana_trazo_auxiliar.htm

Thanks

https://drive.google.com/file/d/0B6oYedIeLTUKczBPY0dfdWdsN0k/view?usp=sharing

ReplyDeleteI think this is different from the previous solutions.

https://www.youtube.com/watch?v=FLLNvVRnUYI

ReplyDelete