See complete Problem 5 at:

www.gogeometry.com/problem/problem005.htm

Triangle, Angles, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Sunday, May 18, 2008

### Elearn Geometry Problem 5

Labels:
angle,
congruence,
triangle

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Triangle CDB is isosceles since BD=DC, therefore angle BCD=angle DBC=2x.

ReplyDeleteLet point E be a point inside ABD triangle so that AED triangle is congruent to CDB triangle, where AE=ED=BD=DC and as it's given AD=BC. Note that both angles EAB and EAD are equal 2x.

Angle ADB = 4x=>angle EDB=2x. Triangle EDB is isosceles since ED=BD.

Angle BAE = angle BAD - angle EAD = 3x - 2x = x.

Let's reflect point E over AB calling it F. Triangles AFB and AEB are congruent. AF=AE, FB=EB. Angle FAE = 2 * angle BAE = 2x.

FAE and BDE are congruent isosceles triangles since AF=AE=DE=DB and angle FAE=angle EDB=2x

=>FE=EB, but FB=EB => triangle FBE is equilateral => angle FBE=60=>AEB=30.

Angle EBD = (180 - angle EDB) / 2 = 90 - x.

Angle ABC = 180 - angle BAC - angle BCA = 180 - 5x.

But angle ABC = angle ABE + angle EBD + angle DBC, or

180 - 5x = 30 + (90 - x) + 2x=> 180 - 5x = 120 +x => x=10

"Note that both angles EAB and EAD are equal 2x"

DeleteWhat? How come? If you subtract <EAD=2x from <BAD=3x, the remainder <BAE=x, not 2x.

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

ReplyDeleteVideo solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 5. Thanks Eder and Cristian.

ReplyDeleteDraw an angular bisector of BDA.

ReplyDeleteDraw DAE=2x, so that the angular bisector of BDA meets it at E.

So, AE=ED (EAD=EDA=2x)

But ∆EAD and ∆BDC are congruent under A.A.S case)

So, ED=BD

Now, Draw a Circle having the center D and Radius DC

Obviously it passes through E, B and C

Let AC and the circle meet at G

So, GD=BD (Radius)

Now connect BG

H be the point of intersection of BG and DE

Since ∆BDG is an Isosceles DH is perpendicular to BG

So, HGD=90-2x

Let F be the reflex of E over AB

As AF=AE, FAB=EAB=x, AB=AB

∆AFB and ∆ AEB are congruent

So FB=EB

Also, ∆AFE and ∆BDE are congruent (BD=AF, BDE=FAE=2x, DE=AE)

So, BE=FE

Eventually EB=EF=BF

So, ∆BFE is an equilateral

Therefore FBE=600

Since FE and AB are perpendicular to each other

ABE=300

Since, Ang.ABC=ABE+EBD+DBC,

ABC=300+900-x+2x

ABC=1200+x

But GBC=90Deg as it is an angle on semicircle (Note that GC is a diameter)

So, ABG=120+x-90

ABG=30+x

Since BGD=BAG+ABG

900-2x=3x+300+x

600=6x

So, x=100

Note: Since ABD = 120-x the proof for the “PROBLEM 4 is inbuilt”

Read "x = 10 degrees" instead "x=100"

DeleteExtend BD to E such that BD = DE

ReplyDeleteVerify BCE is a right angle; Angle BDA = 4x; ABD = Pi - 7x

Now cos 2x = BC / BE = BC / 2BD

=> 2 cos 2x = BC / BD = AD / BD = sin (pi - 7x) / sin3x = sin 7x / sin 3x

=> 2 sin 3x cos 2x = sin 7x

=> sin 5x + sin x = sin 7x

=> sin 7x - sin 5x = sin x

=> 2 cos 6x sin x = sin x

=> cos 6x = 1/2

=> 6x = pi/3

=> x = pi/18 = 10 deg

Dear Antonio - is there a proof using properties of 80-20-80 triangles?

ReplyDeletehttps://www.youtube.com/watch?v=MeEZakv8WxI

ReplyDeleteSince BDC is isosceles, m(BDA) = 4x

ReplyDeleteNow construct an isosceles triangle AED such that m(EAD) = m(EDA) = 2x

the traingles AED and BDE are congruent implying AE = ED = DB and M(BDE) = 2*m(BAE)

=> on careful observation the quadrilateral BAED is similar to problem -4

so m(ABD) = 120-x

from triangle, ABC, 3x+120-x+2x+2x = 180

=> x = 10

Solution:

ReplyDeletehttps://drive.google.com/file/d/0B5CPfrOPKJ1Tem42ZTRZckhjTDQ/view?usp=sharing