## Monday, May 19, 2008

### Elearn Geometry Problem 45

See complete Problem 45
Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

1. Faltan datos, la repuesta asume que AC = CE.. y no se puede llegar a esa conclusion o en todo caso no explica como llega a ese punto.
Espero respuestas.
Muchas Gracias.

2. http://img842.imageshack.us/img842/3018/p045angletriangle.gif

3. Lets use a and b for alpha and beta. BD meets AC on E. Ang(ADE) = 2a+3b and ang(CDE) = a+b. Using sinus rule in ABD and BCD, we get
AB/BD = sin (2a+3b) / sin (2a)
and BC/BD = sin (a+b) / sin (a),
so sin (2a+3b) / sin (2a) = sin (a+b) / sin (a).
That equality gives
sin (2a+3b) = 2.sin (a+b).cos (a)
sin (2a+3b) = sin (2a+b) + sin (b)
sin (2a+3b) - sin (2a+b) = sin (b)
2.sin (b).cos (2a+2b) = sin (b)
and cos (2a+2b)=1/2 or 2a+2b = 60º.
Finally, x+2a = 90º-2b, x = 90º - (2a+2b) = 30º.

4. Typos corrected (regret inconvenience)
Write a for alpha and b for beta.
Draw BE the ⊥ bisector of AC, meeting AD at F.
BE bisects ∠ABC(=4b)
So ∠FBC = 2b,
which implies BD bisects ∠FBC with ∠FBD = ∠DBC = b
Next ∆FAC is isosceles.
So ∠FCA = x, and
∠FCB = ∠FAB = 2a by symmetry
which implies ∠FCD = ∠DCB = a
Follows D is the incentre of ∆BFC, FD bisects ∠BFC.
∠BFD = ∠DFC = x + x = 2x
Denote the sum a + b by y.
From ∆BFC we have 4x + 2a + 2b = 180° and
from ∆ABC we have 2x + 4a + 4b = 180°
2x + y = 90° and x + 2y = 90°
solving which we get x = y = 30°

6. Figure for e-learn geometry problem 45 to follow my proof
http://imageshack.us/content_round.php?page=done&l=img198/5108/elearngeometryproblem45.png#

7. Denote alpha by @ and beta by £

Let the bisector of < ABC meet AD at E. Now D is easily the incentre of Tr. BCE and so < BED = < CED = x.

Since < BAC = 2@+x = < ACB, we have 2x + 4@ + 4£ = 180 .....(1)

Since BE is perpendicular to AC,

x + 2@ + 2£ = 90.....(2)

From (1) and 2 we deduce that
@ + £ = 30 and so from (1) x = 30

Sumith Peiris
Moratuwa
Sri Lanka