## Monday, May 19, 2008

### Elearn Geometry Problem 42

See complete Problem 42
Angles, Triangle. Level: High School, SAT Prep, College geometry

1. draw bisector of ang C
draw ang DAE = α, E on bisector of C
join B to D, D to E, draw altitude BH, H on DE

tr BDE is equilateral
ang DBH = ang DAC ( perpendicular sides )
ang KDE = ang DAC ( AD meetBE on K, AD = EC => DE//AC)
ang FED = ang ECA (CE meet BD on F, DE//AC )

BH bisector of DBE, DK bisector of EDB, EF of DEB
=> DBE = 6 ∙ DAC = 180º
=> DAC = 30º

from tr ABC
A + C + B =( α + 30 ) + ( 30 + α ) + x = 180

x = 120° - 2α
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2. Sir, Did u receive my solution for Problem 42?

1. Yes, Thanks. Following your solution.

3. Draw a circle having its center B and radius BA,
Draw another circle having its center C and radius CB.
Let K be the lower intersection point of two circles.
Connect BK and CK
So, Tri. BKC is an equlateral.
Obviously ang.KBC=60deg
Since ang.BCK=60deg, ang.DCK=60-2α.
Since, ang.DCK is angle on center and ang.DBK is angle on circumferance subtended by same segment: ang.DBK=30-α
similarly BCD=2.BKD
ang.BKD=α
now connect BD
as per the symetrical property in triangles ADB and KBD, ang.ABD also 30-α
so, ang.ABC=X=ang.ABD+ang.KBD+ang.KBC
therefore X=30-α+30-α+60
ie: X=120-2α
............

4. Drop a perpendicular from B to AD say. BE and from C to BD say CF. Tr.s DFC, BFC and ABE are all congruent ASA. So DF = BF = BE, hence Tr. BDF is 30 60 90 with < BDE = 30.

So < ABD = 30-@ and since < FBC =90-@, x = 120-2@

Sumith Peiris
Moratuwa
Sri Lanka

5. Hello, I have seen a very similar problem: B and C are known angles (ie 110 and 20), the angle to find is the one in A position.
Never could solve that problem, I think now it will possible for me if I try to follow the comments here.
Thank you!