Geometry Problem

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

Click the figure below to see the complete problem 685.

## Friday, November 4, 2011

### Problem 685: Isosceles Triangle, 100-40-40 Degrees, Angles, Congruence

Labels:
100 degrees,
40 degrees,
congruence,
isosceles,
triangle

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Awaiting a synthetic Proof:

ReplyDeleteLet AB = BC = a and let AC = b

Given AC = BC + AD

b = a + AD, AD = b - a.

Note BD = AB + AD = BC + AD = AC = b

Apply Sine Rule in triangle ABC:

a/sin 40 = b/sin 100

= b/sin 80 = b/2sin40cos40

So b = 2acos40

Apply Sine Rule in triangle BDC:

a/sinx = b/sin(80-x)

So b sinx = a(sin80cosx-cos80sinx)

2acos40sinx = a(sin80cosx-cos80sinx)

tanx = sin80/(2cos40+cos80)

Problem 685 - To Pravin: and your answer is?

ReplyDeleteThe answer is 30°, but I can't prove it yet, not that easy at first glance, haha!

ReplyDeletex = 30 deg ?

ReplyDeleteUnable to derive it from from above expression for tanx

cirumcenter of Tri ADC

ReplyDeleteAB=a

ReplyDeleteAC=2acos40

Let point E on AC, and CE=AB

AD=AE ---> ang ADE= ang AED=20

AE=AD=a(2cos40-1)=a(4cos^2(20)-3)

DE=2AEcos20=2a(4cos^3(20)-3cos(20))=2a(cos60)=a

---> tri EDC=isosceles ---> ang DEC=160 ---> ang EDC=10

ang ADC=ang ADE + ang EDC = 30

Locate point F on AC such that AD=AF,join B & D to F.We've /_ADF=/_AFD=20 and /_BFC=/_FBC=70. Now, DF/sin(40)=AF/sin(20) or DF=2AFcos(20)-----(1). Further, FC/BF=sin(70)/sin(40) and BF/AF=sin(40)/sin(30);hence FC=AFsin(40)*sin(70)/[sin30)sin40)]=2AFsin(70)=2AFcos(20)-----(2). By (1) & (2), DF=FC but /_DFC=160 hence FDC=10 and thus /_ADC = 20 + 10 = 30

ReplyDeleteHello Pravin,

ReplyDeletesin80=cos10

2cos40+cos80=2cos40+2(cos40)^2-1

2(2(cos20)^2-1)+2((2(cos20)^2-1)^2)-1

=4(cos20)^2-2+8(cos20)^4-8(cos20)^2+2-1

=8(cos20)^4-4(cos20)^2-1

=2cos20(4(cos20)^3-3cos20+cos20)-1

=2cos20(cos60+cos20)-1

=cos20(1+2cos20)-1

=(2cos20-1)(cos20+1)

=(4(cos10)^2-3)(2(cos10)^2)

=(4(cos10)^3-3cos10)(2cos10)

=2cos30cos10

sin80/(2cos40+cos80)

=cos10/(2cos30cos10)=1/(2cos30)=1/root3

tanx=1/root3 ---> x=30

There's yet another way of determining /_ADC which is to locate F on AC with AF=AD as b4. Now let /_FDC=z. Using Ceva's Theorem in trigonometric form: [sin(20)/sin(z)][sin(20-z)/sin(40)][sin(70)/sin(30)]= 1 or sin(20-z)=sin(z) on simplification which gives 20 - z = z or z = 10 hence /_ADC = 20 + z = 20 + 10 = 20

ReplyDeleteLet E be a point on AC so that AE=AD, hence EC=BC. Connect DE, BE and CD.

ReplyDeleteCostruct the altitude CH of the isosceles triangle BCE.

Also the triangle ADE is isosceles.

Because angle A = angle C = 40, then the angle ECH = 20 and the angle EBC=70.

The angle ABE=30 and the angle ADE=AED=20.

In right triangle ECH we have EH = ½BE = EC sin20.

Apply Sine Rule in triangle BDE:

DE/sin30 = BE/sin20 →

DE=½BE/sin20 = EC sin20/sin20 = EC.

Hence DE=EC, then also the triangle DEC is isosceles,

angle EDC=10 and angle ADC=x=20+10=30.

http://www.imagengratis.org/images/p685.png

ReplyDeleteMIGUE.

Κατασκευάζω ισόπλευρο τρίγωνο CDZ στο ημιεπίπεδο

ReplyDeleteAC,B και παίρνω πάνω στη πλευρά ΑΒ ευθύγραμμο τμήμα ΑΕ=ΑD. Παρατηρώ ότι,CBZ=ABC ,EBZ ισόπλευρο,

EDZ,ΕDB ισοσκελή και εύκολα φαίνεται ότι η γωνία

ADB=30 μοίρες

Extend AB and let X and Y be a points on line AB so that x=30°.

ReplyDeleteWhat happenned with my solution o.O?

ReplyDeleteTo Eder: Problem 685. Please send again your solution as I have received only one line.

ReplyDeleteI am Ankur Guria, (brother of Ankan Guria) class 8 from India with a purely geometrical solution. My email i.d. is ankanguria.ju@gmail.com

ReplyDeleteSOLUTION: Cut BC=CE on AC.

So, left AD=AE.

So, in triangle ADE, angles ADE and AED are equal to 20deg. Bisector of Angle BCA is drawn, meeting AB at F. Therefore triangles FBC and FEC are congruent. So, angle EFC = angle BFC = 60deg. Triangles DFE and EFC are congruent. Thus DE=EC. Angle DEC=160deg. So, in triangle DEC, angles EDC and ECD are equal to 10deg. Thus, angle ADC = angle ADE + angle EDC = 20 + 10 = 30deg= x

Excellent Ankur

ReplyDeleteSolution 1

ReplyDeleteForming equilateral triangle BCE internally the triangle ABC, then triangle DBE=triangle ABC (DB=AC ,BE=AB ,<DBE=100-60=40=<BAC). Therefore DE=BC=AB=BE so E is circumcenter BCD so

<BDC=<BEC/2=60/2=30.

Forming equilateral triangle BDE (interior triangle BCD). Then triangle ABC=triangle BCE (BE=AC, AB=BC,<CAB=40=<EBC).So BC=CE, but DB=DE then triangle DBC=triangle DEC.

ReplyDeleteTherefore <BDC=60/2=30.

Solution 3

ReplyDeleteForming equilateral triangle BDE (interior triangle BCD). Then triangle ABC=triangle BCE (BE=AC, AB=BC,<CAB=40=<EBC).So BC=CE, but DB=DE then triangle DBC=triangle DEC.

Therefore <BDC=60/2=30.