## Thursday, May 2, 2013

### Problem 873: Quadrilateral, Diagonal, Triangle, Angle Bisector, Congruence

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

Click the figure below to see the complete problem 873.

1. Reflect BD along AD to B'D
Since AD is the angle bisector, A-C-B' collinear

As DB = DC = DB', D is the center of circle(BCB')
So angleBDB' = 2*angleBCA = 96.

AngleBDA = AngleB'DA = (1/2)*AngleBDB' = (1/2)*96 = 48 = angleBCA
Hence B,D,C,A concylic.

AngleBAC + angleBDC = 180
(5x + 5x) + (48 + x) = 180
x = 12

1. http://img13.imageshack.us/img13/2826/problem873.png

2. There is another solution, namely x=42/5, which occurs when the reflection of BD is CD.

3. http://img23.imageshack.us/img23/8498/problem8731.png

Let circumcircle of ABC cut AD at D’
We have D’B=D’C ( chords of congruent angles) and DB=DC
So DD’ is a perpendicular bisector of BC or D coincide to D’.
To satisfy requirements of the problem, DD’ cannot be a perpendicular bisector of BC So D coincide to D’
Quadrilateral ABDC is cyclic so 48+5x+5x+x=180 => x=12

4. Is there any other way?

5. Yes there is another way

Draw perpendiculars DM & DN to AB & AC which can easily be proved to be equal. Hence Tr. BMD & Tr. CDN are congruent. So < MBD = < DCN = 6x which shows that ABCD is cyclic. Hence < BCD = 5x and adding the angles in Tr.,ACD which add upto 180 and solving we get x = 12

Sumith Peiris
Moratuwa
Sri Lanka

6. Problem 873
The point D is the intersection of the perpendicular bisector BC and bisector <CAB. So by the theorem of South Pole the ABDC is cyclic, then <ABC=x,<DCB=<DAB=5x=<BCD.But
<ABD+<ACD=180 or x+5x+48=5x=180 or x=12.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE