## Sunday, July 6, 2014

### Geometry Problem 1025: Triangle, Double, Triple, Angle, Interior Point

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

Click the figure below to see the complete problem 1025.

1. This problem may be solved in the next way:
1.E be a point on AD such as DE = DC -> Angles: DEC = DCE = 2x
2.Easy to see that DEC is congruent to BDC (SAS) -> EC = BC
3.Angles: ECA = EAC = x = 3x - 2x -> EC = BC = AE
4.Angles: BFE = 180 - BDE - BAD - EBD = 180 - 2x - 8x - (90 + 4x) = 90 -6x
5.Let CG be perpendicular to EB thus EG = BG (Triangle EBC is isoceles)
6.Let EF be perpendicular to AB
7.Triangle AFE ~= BCG ~= GCE (ASA) From simple look we get that in triangle BFE: EF = EG = BG -> Angle
BFE = 30 = 90 - 6x (As we proved above)
Finally x = 10.

2. Let AD meet BC at E and CD meet AB at F. ACEF is clearly cyclic from which we can easily show FE bisects < BED = 6x and FD bisects < ADB = 8x.

So F is the excentre of Tr. BDE and so FB bisects the exterior < . So < ABC must be 8x and solving for x we get 8x + 2x + 5x + 3x = 180 from Tr. ABC and so x = 10

Sumith Peiris
Moratuwa
Sri Lanka