See complete Problem 210 at:
gogeometry.com/problem/p210_triangle_angles_auxiliary_line.htm
Triangle, Angles, Auxiliary Lines. Level: High School, SAT Prep, College geometry
Sunday, November 23, 2008
Elearn Geometry Problem 210: Triangle, Angles, Auxiliary lines
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the solve 30
ReplyDelete[img]http://www.al3ez.net/upload/d/ibrahim_ibrahim_page4.jpg[/img]
The angle between the auxiliary line BE and the segment BD is 30º. Does the data is enought o solve this problem ??
ReplyDeleteOkay Antonio. Thanks, I could to sove this problem. So, All data are correct !
ReplyDeleteangle DBC + angle DEC = 180º. So, quadrilateral BCDE is inscribed. Now, the angle between the auxiliary line BE and the segment BD is 30º. Then, the angle DBE = DCE = a = 30°. On the other hand, angle EDC(=X)is equal to EBC = 90 -2a. So, X= 90 - 2(30)= 30º.
Bolchoi(from Brazil)
Dear bolchoi,
ReplyDeleteThank you for your comments.
My answer:
1. Why angle DBC + angle DEC = 180º?
2. You don't need more data to solve this problem.
3. In fact, you can deduce (it's not a data) that the angle between BE and BD measured 30 degrees.
Dear Jandir,
ReplyDeleteThank you for your comments.
My answer:
1. Why angle DBC + angle DEC = 180º?
2. You don't need more data to solve this problem.
3. In fact, you can deduce (it's not a data) that the angle between BE and BD measured 30 degrees.
Dear Antonio,
ReplyDeleteIn fact I can`t to say at first that angle DBC + angle DEC = 180º. So, my answear at first should be wrong. I`l search better.
Firstly, I trace the line BE. Because the triangle BCE is isosceles(CB = CE), I can deduce that angle DBE = 30º. Now, Let M be a point on the midle point of segment BE(MB = ME)and let N be a point on BD. Then, CM is the height of tringle BCE. Now, because I can deduce that the angle MEN = 30º = DBE, I see that the triangle BNE is isoceles(NB = NE). Now I can deduce that the points B, C, E and N delimit the quadrilateral BCEN which is an inscribed quadrilateral inside an circle. As the points C, M and N are on the same line and the CM is the height of the isosceles triangle BCE, the line CN is the diameter of this circle. Then, the angle NBC = 90º and ABC is a right(or pytagorean) triangle. So, 120 - 2alpha = 90º and then alpha = 15º. Then, I can easily deduce that DBC is an right triangle(BC = BD and angle BDC = angle BCD =45º)and that BCE is an equilateral triangle. Now, I can see that triangle BDE is isosceles(BD = BE). As the angle DBE = 30º, then the angle EDB = 75º. Finally, the measure of the angle CDE = angle EDB - angle BDC = 75º - 45º = 30º
ReplyDeletewhy #BCEN is an inscribed quadrilateral?
ReplyDeleteFrom where did you obtain BCEN is an inscribed quadrilateral ? , and why did you say "let N be a point on BD" , and afterwards you defined it with the angle MEN = 30º ? ok , let's say no problem there but , I repeat from W(here)TF did you obtain BCEN is an inscribed quadrilateral , it's an particular case , i think the real answer is : 2*ALPHA .
ReplyDeleteCheck other cases and it's not 30 degrees .
http://geometri-problemleri.blogspot.com/2009/10/problem-39-ve-cozumu.html
ReplyDeletefie N pe BD astfel iancat CN=a.Se obtine triunghiul NEC echilateral, triunghiul DNC isoscel, triunghiul DNE isoscel si se calculeaza unghiul EDC=30.
ReplyDeleteAntonio,
ReplyDeleteIf alpha=15 deg. then clearly x=30 deg. and this can be done w/o trigonometry.
However, in the general case, is x =2*(alpha)? And how may we prove it?
Please enlighten us.
A²
Dear Ajit, Thanks for your comment. In general case x = 30 degrees. Hint for a geometric prove: draw CF = CB (F on AB extended)
DeleteThanks Antonio.
DeleteGot it.
Problem 210
ReplyDeleteLet point P in the extension of AB to B such that CB=CP=CE (if α<15).Τhen 15.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Let X be midpoint of BE. Let the perpendicular to AB drawn from E meet CX extended at Y and AB at Z.
ReplyDeleteSince BXZY is concyclic and < ZBX = 30 (since < CBE = 90- 2@), < XYZ = < XBY = 30 and Tr.s EXZ and BEY are both equilateral. Also BX = EX = XZ = EZ = ZY
So DE = DY. Now since EY is perpendicular to DB, EY cannot be perpendicular to CD so mark P on CY such that EP is perpendicular to CD.
This yields DE = DP = DY and < DYP = < DPY = DEA, hence CEDY is concyclic
In turn x = < CYE = 30
Sumith Peiris
Moratuwa
Sri Lanka
See graph here.
ReplyDeleteBCE isoscele in C and ∠C = 4.�� ⇒ ∠DBE=30°.
The perpendicular to AB from E intersects AB in F and the angle bisector of ∠BCE in G
The perpendicular to CD from E intersects CD in H and CG in I.
Now ∠EHD = ∠EFD = 90° ⇒ F and H are on the circle of diameter ED ⇒ ∠EDH = ∠EFH.
F middle of GE (easily seen since ∠BEF=60°) and H middle of EI ⇒ FH // GI and ∠EFH = ∠EGI.
So ∠EDC = ∠EDH = ∠EFH = ∠EGI = 30° => x = 30°.
With a link that works to see the diagram.
DeleteAnother solution : see graph here.
ReplyDeleteLet F on AB such that CF = CB = a.
∆BCF isoscele in C ⇒ ∠BCF = 4.α– 60° ⇒ ∠FCE = 60° ⇒ ∆FCE is equilateral.
In ∆CDF, ∠CDF = 60°- α = ∠CDF ⇒ ∆CDF isoscele in F ⇒ FD = FC = FE = a.
So F is the center of the circumcircle to ∆CDE and ∠CFE/2 = ∠CDE = x = 30°
With a new link that works to access the diagram.
DeleteIn the above solution, if α < 15°, ∠ABC is obtuse, F lies on AB extended and segment CB lies within ∠ECF.
ReplyDeleteIn such case, ∠BCF = 60°- 4.α ⇒ ∠FCE = 60° and the remaining of the proof stays the same.
x = 30°
[For easy typing, I use b instead of alpha]
ReplyDeleteConsider Triangle BDC
<BDC=60-b
sin(120-2b)/DC=sin(60-b)/a
DC/a=2cos(60-b)---------(1)
Consider Triangle DEC
<DEC=180-x-b
sin(180-x-b)/DC=sinx/a
DC/a=sin(x+b)/sinx---------(2)
By equating (1) and (2)
2cos(60-b)sinx=sin(x+b)
sin(x+60-b)+sin(x+b-60)=sin(x+b)
sin(x+60-b)=sin(x+b)-sin(x+b-60)
sin(x+60-b)=2cos(x+b-30)sin30
sin(x+60-b)=cos(x+b-30)
sin(x+60-b)=sin(120-x-b)
x+60-b=120-x-b
x=30