Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 739 details.

## Thursday, April 5, 2012

### Problem 739: Triangle, Double Angle, External Angle Bisector

Labels:
angle bisector,
double angle,
external,
isosceles,
triangle

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http://img580.imageshack.us/img580/1161/problem739.png

ReplyDeleteExtend BC to E such that BE=BA ( see sketch)

Note that ∠ (ABD)= ∠ (EBD)….. ( vertical angles and bisector)

So ∆ (ABD)= ∆EBD…. (Case SAS) => ∠ (BED)= ∠ (BAD)= alpha

In Triangle ECD, external angle ∠ACE= 2alpha= ∠ (CED)+ ∠ (CDE) => ∠CDE= alpha

Triangle ECD is isosceles => CD=BC+BE=BC+AB

Locate E on CD such that CE = CB

ReplyDeleteLet F be any point on extn of AB

∠CBE = ∠CEB = α

∠D =∠ FBD - ∠A = 3α/2 - α = α/2

∠EBD = ∠AEB - ∠D = α - α/2 = α/2

∆BCE, ∆ABE, & ∆BED are all isosceles.

Hence CD = CE + ED = BC + BE = BC + AB

let x = AB. Then BC = x/(2cosα).

ReplyDeleteThen AB + BC = x(1 + 1/(2cosα)).

Now consider triangle BCD. By sine law, CD = ((sin(3α/2))*(x/(2cosα)))/sin(α/2).

Hence it is equivalent to prove (sin(3α/2)/sin(α/2) = 2cosα + 1

By considering the double angle formula, 2cosα + 1 = 3 - 4(sin(α/2))^2

The proof is then obvious by the triple angle formula.

Extend AB from B till E so that BE = BC. Draw ED and EC.

ReplyDeleteThe triangle BCE is isosceles then the bisector BD is perpendicular to EC.

Therefore the triangle CED is isosceles, ED = CD.

Let the angle (CBD) = (DBE) = β. Then 2β = 3α.

In the triangle BCD the angle (BDC) = 180 – (180 - 2α) – β = 2α – β; then the angle (EDC) = 2 (BDC) = 4α - 2β = 4α - 3α = α.

Therefore the triangle AED is isosceles. Then AE = AB + BC = ED = CD

Let X be the ex-center of ∆ABC opposite the vertex A.

ReplyDeleteDraw XY ⊥ AB(extended) and XZ ⊥ AD

It is well-known that

the length of the tangent from A to the ex-circle

= AZ = semi-perimeter of ∆ABC

Next ∠XAD = α/2

and ∠XDA = ∠YBD - ∠A = 3α/2 - α = α/2

So ∆AXD is isosceles and XZ ⊥ AD

Follows Z is the mid-point of AD

∴ AC + CD = AD = 2AZ = Perimeter of ∆ABC = AC + AB + BC,

Hence CD = AB + BC

The solution is uploaded to the following link:

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJN0FmNktoVHF5Vlk

To Rengaraj

ReplyDeleteYour sketch and the presentation is really nice.

I wonder what software did you use to draw it .

Please let us know.

I use corelDRAW X5 for sketching, then I export the diagram as JPG to Microsoft word 2010. I have a software called MathType to make efective mathematic typing.

ReplyDeleteThaks Peter

Have a nice Day.

Picture: http://i1237.photobucket.com/albums/ff480/Evan_Liang/DoubleangleProblem1.jpg

ReplyDelete∠BAC=a ∠BCA=2a

Locate F on AC so that BF=AF

∴∠BAC=∠ABF=a (base angles theorem)

∴∠BFC=2a (exterior angles theorem)

∵∠BCA=2a

∴BF=BC (converse of base angles theorem)

locate E on the extended segment of AC so that AE=AB

∴∠BEA=∠EBA=a/2

Goal: Prove CE=FD so then EF=CD=AE+AF=AB+BC

∵∠GBC is the exterior angle of ∠ABC

∴∠GBC=3a

∴∠CBD=3a/2

∵∠EBA=a/2,∠ABF=a

∴∠EBF=a+a/2=3a/2=∠CBD

∴∠EBC=∠FBD

∵BC=BF,∠BFC=∠BCF

∴△CBE≅△FBD

∴CE=DF

∴CD=EF=AE+AF=AB+BC

Solution 739 sent by Sumith Peiris, Sri Lanka

ReplyDeleteFind E on CD such that BE=ED

Now < CBD=3A/2 and so <BDE=<BED=A/2 (A= alpha)

So Tr BCE and Tr ABE are also isoceles, hence CD= CE+ED= BC+AB

Sumith Peiris

Sri Lanka