See complete Problem 184 at:
www.gogeometry.com/problem/p184_triangle_angles_problem.htm
Triangle and angles. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Friday, September 26, 2008
Elearn Geometry Problem 184
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x=32
ReplyDeleteEvrenin, the answer x = 32 is not correct, try again. Good luck.
ReplyDeletex=30. I have an algebraic proof. Antonio, do you have a geometric one?
ReplyDeleteOrange Skid
Yes, I have a geometric solution.
ReplyDeleteThank you for your question.
Angle ABD=58 which makes angle ABC = 88 and angle ACB =44. AC/AD=sin(x+16)/sin(x) =AC/AB=sin(88)/sin(44) which leads us to a simple equation like: sin(x+16)/sin(x) = 2cos(44) which gives x=30 deg since sin(46)=cos(44). Thus, x=30 deg.
ReplyDeleteI still would like to know how this done by plane geometry alone.
Ajit: ajitathle@gmail.com
Problem 184
ReplyDeleteForming an equilateral triangle ADE (E is above the AC),then AB=AE=AD=ED .But <EAB=4,
<EBA=(180-4)/2=88=58+30, so<EBD=30. Then B, E,C are collinear.But <EDB=2,<DEC=32, so
<EAC=44=<ECA ,so EC=AE.Therefore E is the center of the circle triangle ADC so
<ACD=<AED/2=60/2=30.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Let X be midpoint of BD. Let DZY be perpendicular to BC, Z on BC and Y on AX extended.
ReplyDeleteBXZY is concyclic and hence Tr.s DXZ and BDY are both equilateral.
So BX = DX = XZ = DZ = ZY and hence CD = CY.
Now mark P on AY such that CP = CY (= CD). So < CYP = < CPY = < CDQ where Q is on AD extended.
It follows that ADCY is concyclic and x = < AYD = 30
Sumith Peiris
Moratuwa
Sri Lanka
Sumith: I do not understand "Let DZY be perpendicular to BC, Z on BC and Y on AX extended." Would you kindly elaborate or provide a diagram please.
DeleteLet me rephrase "Let DZY be perpendicular to BC, Z on BC and Y on AX extended."
Delete"Drop a perpendicular from D to BC to meet BC at Z. AX and DZ extended meet at Y"
Trust this clarifies
It is special case of problem (t, 3t, 30) where t = 16.
ReplyDeleteIt is the same problem with following cevian problem (Circumcircle BCD meets AC at P)
https://output.jsbin.com/fofecum#48,16,14,44
Sir Mr.Sumith, Please may I ask; kindly explain how are
ReplyDelete"BXZY is concyclic and hence Tr.s DXZ and BDY are both equilateral".
Thanking you in advance.
That BXZY is concyclic (with BY diameter) is clear because Tr.s ABX & ADX are congruent SSS, <BXZ = 90 = <BZD (by construction)
DeleteNow BDZ is a 30-60-90 triangle so DXZ is equilateral
Now Y lies on the perpendicular bisector of BD and hence BY = DY. But < BDY = 60 so Tr. BDY is equilateral
Trust this clarifies