Proposed Problem

See complete Problem 18 at:

gogeometry.com/problem/problem018.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Saturday, February 21, 2009

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

Labels:
angle,
congruence,
right triangle

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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

ReplyDeleteAjit:ajitathle@gmail.com

It's possible to solve this problem without using trigonometry.

ReplyDeleteThe 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."

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.,

DeleteGiven 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.,

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

ReplyDeletehttp://i1237.photobucket.com/albums/ff480/Evan_Liang/DoubleangleProblem2.jpg

ReplyDeleteA 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.)

∴△ADE∼△AEC (AA similarity)

∴DE/CE=AE/AC ...................(2)

Use transitive Prop. between (1) and (2)

∴DE/CE=2BE/CE

BE=2BE

Produce CB to F such that EB = BF

ReplyDeleteJoin AF

rtΔABF and rtΔABE congruant (SAS)

Angle(FAE)=Angle(EAD)=2α

Angle(AFB)=Angle(AEB)=2α + θ

Agle(EDC)= 2α + θ =Angle(AFB)

ADEF concyclic

DE = EF = 2BE (equal circumference angle equal chord)

Start like the other solutions and reflect ABE to a new triangle ABF.

ReplyDeleteThen 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.

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

ReplyDeleteLet AE=1

ReplyDeletesin(a) = BE/AE

BE=sin(a)

<AEC=90+a

<DEC=90+a-0

<ADE=90+a

By sine law,

sin (90+a) / AE = sin(2a)/DE

cos(a) = 2sin(a)cos(a)/DE

DE=2sin(a)

BE=sin(a) (proved above)

Hence DE=2BE proved.