See complete Problem 9 at:

www.gogeometry.com/problem/problem009.htm

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

Post your solutions or ideas in the comments.

## Sunday, May 18, 2008

### Elearn Geometry Problem 9

Labels:
angle,
congruence,
triangle

Subscribe to:
Post Comments (Atom)

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

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

ReplyDeleteI am brazilian, sorry for the bad english.

ReplyDeleteLet 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.

The solution is uploaded to the following link

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJdW5xcjU2OHVxbGM

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.

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

ReplyDeleteLet 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.

ReplyDeleteBest regards,

Stan Fulger

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.

ReplyDeleteBest regards,

Stan Fulger

Extend BD to E such that BD = DE

ReplyDeleteThen < 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

This comment has been removed by the author.

ReplyDeletehttps://www.youtube.com/watch?v=R3K8SkKsaaw

ReplyDeleteConstruct a line AE such that AE = BC and is at an angle of 30 degree to AD.

ReplyDeleteJoin 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

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.

ReplyDeleteTriangle 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

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

ReplyDeleteFirstly, 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°

Triangle ABD is equalaterial

ReplyDeleteso, angle ABD=angle BAD=60

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

sin10/AD=sinx/CD

AD/CD=sin10/sinx

Since AD=BD

sin40/sin(x+30)=sin10/sinx

sin40/sin10=sin(x+30)/sinx

It is easy to point out that x=10

As <B=90+x, the circumcenter of triangle ABC lies onto AD and since D is also onto perpendicular bisector of AB, D is its circumcenter, so <C=30, <BCD=40=<CBD=x+30, x=10. At this point I cannot see if this solution has already been posted.

ReplyDeleteBest regards