## Saturday, February 21, 2009

### Problem 18: Right triangles, Angles, Congruence

Proposed Problem

See complete Problem 18 at:
gogeometry.com/problem/problem018.htm

Level: High School, SAT Prep, College geometry

1. Let angle theta = t & alpha = a. Since B=90 we've 3a + t = 90 or t = 90-3a. In tr. ADE, DE/AE = sin(2a)/sin(180 -2a-90+3a)=sin(2a)/sin(90+a)=2sin(a)*cos(a)/cos(a)=2sin(a) or DE=2sin(a)*AE. From tr. ABE, BE =sin(a)*AE. Thus, DE=2BE
Ajit:ajitathle@gmail.com

2. It's possible to solve this problem without using trigonometry.
The key is auxiliary construction:

Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

1. Given sufficient data any such problem can be solved using Trigonometry. The beauty is to solve using pure geometry alone which I think is the purpose of this website.,

2. Given sufficient data any such problem can be solved using Trigonometry. The beauty is to solve using pure geometry alone which I think is the purpose of this website.,

3. If you construct EB extending to the same length ending at F so that EB = BF and then join AF. I am not sure if we can show by triangle similarity that EF = DE which means DE =2 BE

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

6. http://i1237.photobucket.com/albums/ff480/Evan_Liang/DoubleangleProblem2.jpg
A much easier way using angle bisector and similarity
construct a reflection of △ABE △ABKalong the line of reflection AB
∴△ABK≅△ABE
∴∠KAB=∠BAE=a,AK=AE,KB=BE
∵∠EAC=2a=∠KAE
∴AE is the angle bisector of ∠KAC
∴KE/AK=CE/AC=2BE/AE
AE/AC=2BE/CE ..................(1)
∵∠AED=∠ECA,∠EAC=∠EAC (reflexive prop.)
∴DE/CE=AE/AC ...................(2)
Use transitive Prop. between (1) and (2)
∴DE/CE=2BE/CE
BE=2BE

7. Produce CB to F such that EB = BF
Join AF
rtΔABF and rtΔABE congruant (SAS)
Angle(AFB)=Angle(AEB)=2α + θ
Agle(EDC)= 2α + θ =Angle(AFB)
DE = EF = 2BE (equal circumference angle equal chord)

8. Start like the other solutions and reflect ABE to a new triangle ABF.
Then add point G on the line AC such that AGE = 90 - a
tri(EDG) is then isosceles and EG = ED. In addition tri(AEG) is isosclese with AE = AG.
So tri(AEG) is congruent to tri(AFE) via SAS and FE = EG = ED = 2BE from that congruence.

9. Nearly same proof as Anonymous above except I proved AECD con cyclic by noting that AF is tangential to Tr. FDC since AE is tangential to Tr. AEC

10. Let AE=1
sin(a) = BE/AE
BE=sin(a)
<AEC=90+a
<DEC=90+a-0