## Sunday, May 18, 2008

### Elearn Geometry Problem 5

See complete Problem 5
Triangle, Angles, Midpoint, 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”

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?

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. Grab a pen and a piece of paper, and illustrate along.

Firstly, let us calculate some angles:
Angle BCD = Line AE = line DE = line BD = line CD
> Triangles ADE and BCD are identical.
> Angle BAE = x
> Angle BDE = 2x

Add point F to line AB and connect it to point D such that line DF is the angle bisector of angle BDE. Now, connect point F to point E.

We can see that:
> Angle BDF = angle EDF = x
> Triangles BDF and EDF are congruent by SAS because:
· Angles BDF and EDF are equal.
· Line BD = line DE
· They share a side.
> Triangle ADF is isosceles because angle DAF = angle ADF = 3x.
> Since triangle ADF is isosceles, line AF = line DF.
> Triangles FAE and EDF are congruent by SAS because:
· Angle FAE = angle FDE = x
· Line AE = line DE
· Line AF = line DF

Let angle AFE have a value of q such that angle AFE = angle DFE = angle DFB = q.

Now, we solve the problem:
> 3q = 180°, angle DFB = q = 180°/3 = 60°
> Since all three angles in a triangle add up to 180°,
180° - 7x + x + 60° = 180°
180° - 6x = 120°
6x = 60°
x = 60°/6 = 10°

11. This comment has been removed by the author.

12. <BCD=2x
<BDA=4x
<ABD=180-7x

Consider triangle and by sine law

Consider triangle BDC and by sine law again
sin(180-4x)/BC=sin2x/CD
BC/CD=sin4x/sin2x=2cos2x--------(2)

Equating (1) and (2) as AD=BC & BD=CD
sin7x=2sin3xcos2x
sin7x=sin5x+sinx
sin7x-sin5x=sinx
2cos6xsinx=sinx
sinx(2cos6x-1)=0
cos6x=1/2
6x=60
x=10