See complete Problem 45

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

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 45

Subscribe to:
Post Comments (Atom)

skip to main |
skip to sidebar
## Monday, May 19, 2008

###
Elearn Geometry Problem 45

## Search This Blog

## Blog Archive

## Link List

Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 45

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

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

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.

ReplyDeleteEspero respuestas.

Muchas Gracias.

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

ReplyDeleteLets 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

ReplyDeleteAB/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.

Typos corrected (regret inconvenience)

ReplyDeleteWrite 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°

http://s1092.photobucket.com/albums/i418/vprasad_nalluri/?action=view¤t=elearngeometryproblem45.jpg

ReplyDeleteFigure for e-learn geometry problem 45 to follow my proof

ReplyDeleteLink:

http://imageshack.us/content_round.php?page=done&l=img198/5108/elearngeometryproblem45.png#

Denote alpha by @ and beta by £

ReplyDeleteLet 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

Hi Sumith.

DeleteNot important, but < BED = < CED = 2x (you made a mistake)

Equations (1) and (2) are the same.

From (1) and (2) ... how to deduce that @ + £ = 30 ?

Pedro

Hi,

DeleteThis is my solution

https://mega.nz/#!UhwWVI4Z!yOpCiF1X1u59Wlg0A293tGYRHKzZ-l7CXOZVQzLooLE

Regards

Pedro Miranda

Dear Pedro, yes u are right, there is a mistake in my proof above

ReplyDeleteAmended proof is as follows:-

Let the bisector of < ABC meet AD at E.

In Isoceles Tr. ABC 2x + 4alpha + 4 beta = 180 .. (1)

Now D is obviously the incentre of Tr. BCE

Hence < BED = 2alpha + 2 beta = < DEC = 2x or x = alpha + beta

Now substituting in (1)

2x + 4x = 180 whence x = 30