## Tuesday, January 5, 2010

### Problem 414: Triangle, Angles, Altitude, Median, Congruence

Proposed Problem
Click the figure below to see the complete problem 414 about Triangle, Angles, Altitude, Median, Congruence.

Complete Problem 414
Level: High School, SAT Prep, College geometry

1. http://geometri-problemleri.blogspot.com/2010/01/problem-61-ve-cozumu.html

2. mark ang ABE as y and BDC as z

draw 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

3. /_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.
Ajit

4. Angle CEB = 66 degrees, see circle with diametr AB - that angle BAC is equal 66/2 = 33 degrees.

5. Joe, yuv-k:
'Angle ABC = 90' is not an original data.

6. c.t.e.o:
If M is midpoint of EB,
angle FEM = angle EFM

7. thanks, I know it now

but at solution of joe angle B is 2x+24 ???

8. Okay I now see my error. I've assumed 66-x = x which has not been proven. Sorry about that.
Ajit

9. in every triangle,at every vertex,the angles (side,atitude),(circumradius,other side) are congruent
centroid 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
.-.

10. To Anonymous:
Why does circumcenter O lie on the line AM?

11. sorry ,i have other point names;
centroid 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

12. bjhvash44@sbcglobal.net

AC 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

13. angle(ABE)=66-x, angle(EBD)=90-x.
angle(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).

14. extend BD to G, BE to H, G and H on circle of ABC
=> arc AH = arc GC => ▲AEH = ▲CEG => GEC = 66° =>
EC bisector => E center => x = 1/2 BEC = 33°

15. In this triangle, altitude and median from B are isogonals.
This only can happen when the triangle is rectangle in B and solution follows

16. it's actually "triangle is right"

17. Video solution
http://youtu.be/3iAHgFDKmUQ

Greetings go-solvers!

18. Let the perpendicular to AC drawn thro’ E meet AB at F.

Then < 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