## Thursday, December 24, 2015

### Geometry Problem 1173: Triangle, Cevian, Angle, Congruence, Similarity

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1173.

1. Consider Tr.DBA and Tr. ABC,
AB/BC = DB/AB = 2/3 Angle B is common, thus Tr.DBA and Tr. ABC are similar hence Angle BAD = Angle C
Lets assume bisector of angle B meets AC at F, AF/FC=2/3. FC=3AC/5 And CE=AC/3 and it is easy to see FE/CE = 4/5, hence BF//DE and it proves Angle ABD = 2 × Angle CDE.

3. http://s15.postimg.org/o5x05aenv/pro_1173.png

Verify that BA^2=BD.BC => BA tangent to circumcircle of tri. ADC
Draw BF//DE
In triangle ABF we have ∠(ABF)= ∠(BFD)- ∠ (BAD)
In triangle DEC we have ∠ (EDC)= ∠ (AED)- ∠ (ACD)
But ∠ (AED)= ∠ (ADE)= ∠ (BFD)
And ∠ (FBD)= ∠ (EDC)
Replace in above expressions we get ∠ (ABD)=2. ∠ (EDC)
Triangles DEC and BFA are similar ..( case AA)
So CE/AF=DC/AB= 5/6
Since BF bisect angle ABD so
AF/AB=FD/BD= > AF= 3/5 x AD

4. In Triangles BAD, BCA we have
BA/BC=6/9=2/3=4/6=BD/BA and angle B is common.
By SAS they are ///.
Also Corresponding anglea BAD, BCA are equal.
Hence Angles

5. AB/BC=6/9=4/6=BD/BA ,therefore tri ABD and tri CBA is similar, then <BAD=<BCA.
Bring from C parallel to the KE intersecting AK to L( CL//KE), then is KL=DC=5,BK=BD=4
AK=10, but AE/EC=AK/KC=10/5=2.Therefore AE=2EC.

6. AB/BC=6/9=4/6=BD/BA ,therefore tri ABD and tri CBA is similar, then <BAD=<BCA.
The ED intersects AB at F . Is <BAD=<ACB=x and <EDC=y, then <ADE=<AED=x+y, but
But <BKC=<BFD=y=<BCK
Bring from C parallel to the DE intersecting AF to K( CK//FE), But <BKC=<BFD=y=<BCK
then is KF=DC=5,BF=BD=4

7. We use Barycentric Coordinates.
Definition: Capital letters will denote vectors. P = x A + y B + z C will be written as P = (x, y, z). By definition, x + y + z = 1. AB means “length of AB” unless otherwise specified; most importantly, it does NOT mean “A times B”.
Let A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1).
As BD : DC = 4 : 5, it is clear that D = (0, 5/9, 4/9).
If P – M = (u, v, w), then PM is given by:
d^2 = -(vw)a^2 – (uw)b^2 – (uv)c^2, where a, b, c are BC, AC, AB respectively.
Plugging in D – A into the distance formula results in AD = 2b/3. It is given that AE = AD, hence E = (1/3, 0, 2/3). It is obvious that (1/3)A + (2/3)C = E, hence AE:EC = 2:1, and therefore AD = 2CE. [One part of the question -> tick!]
Construct G on such that G = (5/9, 0, 4/9). This means that AG : GC = 4 : 5, with G on line AC. This ratio and the fact that C is shared by both Triangle GDC and Triangle ABC indicate that the named triangles are similar and that AB is parallel to GD.
From the similar triangles, AB : GD = BC : DC, so GD = 10/3. It’s trivial to find the following facts: E = (3/5)G + (2/5)C which indicates that EG : EC = 2:3; GD : DC = 2 : 3. By the angle bisector theorem, line DE bisects angle GDC. With parallel lines, Angle GDC = Angle ABC. Hence, twice of Angle EDC = Angle ABC. [Second part of question -> tick!]
The third part of question is simple angle chasing.
Angle AED = Angle CDE + Angle ECD. (External angle, internal opposite angles of Triangle CDE)
Angle AED = Angle ADE (Isosceles triangle)
Hence, Angle ADC = (Angle CDE + Angle ECD) + Angle CDE
Angle ADC = Angle ABD + Angle BAD (External angle, internal opposite angles of Triangle ABD)
Combine two expressions of Angle ADC.
Angle ABD + Angle BAD = (Angle CDE + Angle ECD) + Angle CDE
Therefore, Angle BAD = Angle ECD. [Question done, boys!]