## Friday, November 4, 2011

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

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

Click the figure below to see the complete problem 685.

1. Awaiting a synthetic Proof:
Let AB = BC = a and let AC = b
Given AC = BC + AD
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)

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

4. x = 30 deg ?
Unable to derive it from from above expression for tanx

6. AB=a
AC=2acos40
Let point E on AC, and CE=AB
DE=2AEcos20=2a(4cos^3(20)-3cos(20))=2a(cos60)=a
---> tri EDC=isosceles ---> ang DEC=160 ---> ang EDC=10

7. 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

8. Hello Pravin,

sin80=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

9. 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

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

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

MIGUE.

12. Κατασκευάζω ισόπλευρο τρίγωνο CDZ στο ημιεπίπεδο
AC,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

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

14. What happenned with my solution o.O?

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

16. I 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

SOLUTION: Cut BC=CE on AC.
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

17. Excellent Ankur

18. Solution 1
Forming 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.

19. 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.
Therefore <BDC=60/2=30.

20. Solution 3
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.
Therefore <BDC=60/2=30.