Proposed Problem
Click the figure below to see the complete problem 414 about Triangle, Angles, Altitude, Median, Congruence.
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Complete Problem 414
Level: High School, SAT Prep, College geometry
Tuesday, January 5, 2010
Problem 414: Triangle, Angles, Altitude, Median, Congruence
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mark ang ABE as y and BDC as z
ReplyDeletedraw EF perpendicular to AB => ang FEB = z
FEDB is concyclic, if mark M midpoint of EB =>
ang FEM = ang EMF = z
ang MFB = ang FBM = y
from tr FMB = > z = 2y
from tr BDC = > z = 60, y = 30
from tr ABC = > x = 36
Wrong answer.
Delete/_BED=90-24=66 deg. and thus /_CBD=/_ABE= 66-x. This makes /_BCA=90-(66-x)=24+x. Triangle ABC gives us: (24+2x)+x+(24+x)=180 or x=33 deg.
ReplyDeleteAjit
Angle CEB = 66 degrees, see circle with diametr AB - that angle BAC is equal 66/2 = 33 degrees.
ReplyDeleteJoe, yuv-k:
ReplyDelete'Angle ABC = 90' is not an original data.
c.t.e.o:
ReplyDeleteIf M is midpoint of EB,
angle FEM = angle EFM
thanks, I know it now
ReplyDeletebut at solution of joe angle B is 2x+24 ???
Okay I now see my error. I've assumed 66-x = x which has not been proven. Sorry about that.
ReplyDeleteAjit
in every triangle,at every vertex,the angles (side,atitude),(circumradius,other side) are congruent
ReplyDeletecentroid G,circumcenter O lie on the line AM
AM is the Euler line of triangle ABC
orthocenter H lies on the Euler line AM and on the altitude AD
H=A
BAC is a rigth angle, x=33
.-.
To Anonymous:
ReplyDeleteWhy does circumcenter O lie on the line AM?
sorry ,i have other point names;
ReplyDeletecentroid G, circumcenter O lie on the line BE
BE is the Euler line of triangle ABC
orthocenter H lies on the Euler line BE and on the altitude BD
H=B
ABC is a rigth angle,x=33
to prove the first theorem
forget median BE
draw the circumcircle of ABC and diameter BF
ABF is a rigth triangle
inscribed angles AFB and ACB are congruent
the triangles AFB and BDC are similar
angles CBD and OBA are congruent
that's why in this problem O lies on the line BE
bjhvash44@sbcglobal.net
ReplyDeleteAC is the longest side. A perpendicular @M midpoint passes thru the center of the triangles circumscribed circle so AC is achord of this circle. Only way angle a and a can be equal is when AB is a diameter angle ABC = 90 x=33
angle(ABE)=66-x, angle(EBD)=90-x.
ReplyDeleteangle(BCD)=24+x
by sine law, in tr(ABE),
sinx/sin(66-x)=BE/AE,
and in tr(BEC),
sin(24+x)/sin(90+x)=sin(24+x)/cosx=BE/EC
therefore, because AE=EC
cosx*sinx=sin(66-x)*sin(24+x).
Answer is x=33
extend BD to G, BE to H, G and H on circle of ABC
ReplyDelete=> arc AH = arc GC => ▲AEH = ▲CEG => GEC = 66° =>
EC bisector => E center => x = 1/2 BEC = 33°
In this triangle, altitude and median from B are isogonals.
ReplyDeleteThis only can happen when the triangle is rectangle in B and solution follows
it's actually "triangle is right"
ReplyDeleteVideo solution
ReplyDeletehttp://youtu.be/3iAHgFDKmUQ
Greetings go-solvers!
ok
DeleteLet the perpendicular to AC drawn thro’ E meet AB at F.
ReplyDeleteThen < CFE = < AFE = < ABD = < CBE.
So BCEF is concyclic and so ABC is a righ angled triangle with centre E.
Therefore < BED = 2x and so x = ½(90-24) = 33
Sumith Peiris
Moratuwa
Sri Lanka
<BED=66
ReplyDelete<ABE=66-x=<DBC
<BCD=24+x
sinx/BE=sin(66-x)/AE
BE/AE=sinx/sin(66-x)-------(1)
sin(24+x)/BE=sin(90-x)/CE
BE/CE=sin(24+x)/sin(90-x)---------(2)
Since AE=CE, (1)=(2)
sinx/sin(66-x)=sin(24+x)/cosx
sin(66-x)sin(24+x)=sinxcosx
cos(42-2x)-cos90=sin2x
sin(48+2x)=sin(180-2x)
48+2x=180-2x
x=33