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

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

Click the figure below to see the complete problem 882.

## Saturday, June 1, 2013

### Problem 882: Triangle, Angle, x, 2x, 7x, Cevian, Congruence, Auxiliary Lines

Labels:
angle,
auxiliary line,
cevian,
congruence,
double angle,
triangle

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Ubicamos E en el segmento AC, tal que AD=DE=EC, entonces el ángulo EDB=4x. Luego trazamos el segmento EB y la bisectriz del ángulo ECB que es CF (F en EB), entonces este segmento es parte de la mediatriz del segmento EB. También proyectamos el punto E sobre el segmento AB (punto G) y trazamos el segmento EG, con esto logramos tres triángulos congruentes: DGE, CFE y CFB. Estos tres triángulos congruentes forman el triángulo notable BGE, donde el ángulo GBE mide 30 grados sexagesimales. Así, el ángulo BEC mide 30+2x grados; finalmente en el triángulo EFC tenemos que 30+6x=90 de donde x=10 grados.

ReplyDeleteEnglish please..

DeleteDraw an arc with radius BC from point C intersects AC and AB on point E and F respectively. Triangle BFC and GBC are congruent triangle then angle FBC = angle GCB = 8x.

ReplyDeleteBy the sum of interior angles theorem on triangle ABC:

8x + 8x + 2x = 180

x = 10.

Where is point G located?

Deletethe given property AD=BC is not used.

Deleteby the way, where is point G in your solution?

To Falensius

ReplyDeleteI note that your solution never use the given information AD=BC .Is it true that AD=BC is not require to get x=10 ?

Peter Tran

On line AC the length is equal to BC.

ReplyDeleteI've never used the information AD = BC. I just add more point and lines to solve the problem without considering the information. I will try to find another solution to solve the problem with considering the information AD = BC.

ReplyDeletethank u.

To Falensius. In your solution of problem 882, why triangles BFC and GBC are congruent?

DeleteTrigonometry solution:

ReplyDeletelet put AC=1

in triangle ACD, AD=sin(x)/sin(3x)

in triangle ABC, BC=sin(2x)/sin(10x)

equation AD=BC have 3 solutions, x=0, x=10 and x=45

x=0 correspond to AD=BC=0

x=10 correspond to AD=BC= finite number

x=45 correspond to AD=BC= infinity

See my solution, mentioned below:

ReplyDeletehttp://imageshack.us/f/708/p882.png/

Let CM be the bisector of < ACB, M on AB.

ReplyDeleteLet MN be drawn perpendicular to DC, N on AC

Then Tr. s DMC, BMN, BCN, CND are all isoceles

So Tr.s DMN, CMN and BCM are all congruent SSS.

So < DMN = < CMN = < CMB = 6x so 6x + 6x+ 6x = 180 and x = 10

Sumith Peiris

Moratuwa

Sri Lanka