## Sunday, May 18, 2008

### Elearn Geometry Problem 9

See complete Problem 9 at:
www.gogeometry.com/problem/problem009.htm

Triangle, Angles, Congruence. Level: High School, SAT Prep, College geometry

1. http://geometri-problemleri.blogspot.com/2009/05/problem-22-ve-cozumu.html

2. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

3. I am brazilian, sorry for the bad english.
Let E be the intersection of AC and BD. For triangle AED, angle(AEB) = x + 60. For triangle EBC, angle(ECB) = angle(AEB) – angle(EBC) = x + 60 – (x + 30) = 30.
So angle(ADB) = 2.angle(ACB), points A, B and C belong to a same circle with center D, AD = DC, triangle ADC is isosceles and x = 10.

5. As Nilton has pointed out: angle(ADB) = 2* angle(ACB) AND AD = BD which implies that D lies on the perpendicular bisector of AB which, in turn, means that D is the circumcentre of Tr.ABC; hence etc.

6. To Ajit: You're right, the condition AD = BD should be mentioned explicitly, though it is said in the enunciate. Thanks for your help.

7. Let perpendÄ±cular bÄ±sector of AB (passing through D) intersect AC at E; <ADE=30, so <CED=30+x=<CBD and BEDC is cyclic, so <DBE=<DCE=10. As, by symmetry <DAE=<DBE, our answer is <DAE=10.

Best regards,
Stan Fulger

8. Triangle ABD is equilateral; let the perpendicular bisector of AB intersect AB at E and, clearly <CED=30+x=<CBD, so BEDC is cyclic and <DBE=<DCE=10 ( 1 ). Since by symmetry <DAE=<DBE, we have our answer, x=10.

Best regards,
Stan Fulger

9. Extend BD to E such that BD = DE

Then < EAC = < EBC = x+30 and so ABCE is cyclic with D as centre
Hence Tr. ADC is isoceles and x=10

Sumith Peiris
Moratuwa
Sri Lanka

10. This comment has been removed by the author.

12. Construct a line AE such that AE = BC and is at an angle of 30 degree to AD.
Join DE. The trianlges BAE and ABC are congruent (SAS)
=> m(AEB) = m(ACB) =y (say) => A,E,C,B are concyclic.
=> m(AEC) + m(ABC) = 180
=> m(AEB) + m(BEC) + m(ABE) + m(EBC) = 180
=> y+(60-x)+60+(x+30) = 180
=> y = 30
=> D is the center of the circle passing through A,E,C and B
Now consider the isosceles triangle BDC in which BD = BC
=> x+30 = 40
=> x = 10 degrees

13. Construct point E such that DE is the perpendicular bisector of triangle BCD, such that <BED=90. Construct point F, which is the intersection point of ED and AC. Join BF.

Triangle EFB is congruent to EFC (SAS). Since <BCA=30, <EBF=30.
Hence <FBD=x. Also, BF=CF.
Since BF=CF, FD=FD, <DFB=<DFC,
triangle DFB is congruent to DFC (SAS).
Hence <FBD=FCD,
x=10

14. Grab a pen and a piece of paper, and illustrate accordingly.

Firstly, let us calculate some angles:
> Angle BAC = 60° - x
> Angle ABC = 90° + x

Add point E to the figure and connect it to points B and C such that lines BC, CE and BE are equal. The newly constructed equilateral triangle BCE should overlap quadrilateral ABCD. Next, connect point E to point A. Label the intersection point of lines BE and AC as point F.

We can conclude that:
> Angle ACE = 60° - 30° = 30° = Angle ACB
> Line AC bisects angle BCE.
> The angle bisector theorem states that (BF/EF) = (BC/CE). Since BC and CE are equal, BF and EF are equal.
> Angle BFC = 180° - 60° - 30° = 90°
> Angle BFC = angle CFE = angle EFA = angle AFB = 90°

Now, we solve the problem:
> Triangles ABF and AEF are congruent by SAS because:
· Line BF = line EF
· Angle AFE = angle AFB
· They share side AF.
> Since triangles ABF and AEF are congruent, line AB = line AE.
> Since lines AE and AB are equal, triangle ABE is isosceles.
> Triangles ABE and BCD are congruent by SAS because:
· Line AB = line BD
· Angle ABE = 90° + x - 60° = 30° + x = Angle CBD
· Line BE = line BC
> So, triangle BCD is isosceles because triangle ABE is isosceles.
> Angles CBD and BCD are equal because triangle BCD is isosceles.

x + 30° = 40°
x = 40° - 30° = 10°

15. Triangle ABD is equalaterial
angle BAC=60-x
angle BEC=120-x (E is the intersecting point of BD and AC)
angle BCA=30
angle BCD=30+10=40

Consider triangle BCD
sin40/BD=sin(x+30)/CD
BD/CD=sin40/sin(x+30)

Consider triangle DAC