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