## Sunday, May 18, 2008

### Elearn Geometry Problem 5 See complete Problem 5 at:
www.gogeometry.com/problem/problem005.htm

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

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

Let 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

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

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

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

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

4. Draw an angular bisector of BDA.
Draw DAE=2x, so that the angular bisector of BDA meets it at E.
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
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”

1. 5. Extend BD to E such that BD = DE
Verify 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

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

7. 8. Since BDC is isosceles, m(BDA) = 4x
Now 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

9. Solution:
10. 