Geometry Problem
Click the figure below to see the complete problem 505 about Right Triangle, Cevian, Sum of Segments, Angles.
See also:
Complete Problem 505
Level: High School, SAT Prep, College geometry
Wednesday, August 18, 2010
Problem 505: Right Triangle, Cevian, Sum of Segments, Angles
See also:
Complete Problem 505
Level: High School, SAT Prep, College geometry
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With the given data is it possible to prove that BD bisects /_B ? Easy to do this by trigonometry but how do we do it by plane geometry alone?
ReplyDeleteOf course, this makes x=15 deg.
Vihaan
draw DBM = x => BM = a, get AH = a + d , H on AC
ReplyDelete=> CH = DM => HM = d => MA = a => MBA = 2x, B = 6x
=> x = 15°
https://photos.app.goo.gl/BkkquzcTCvWzbeUP6
DeleteTo c.t.e.o
ReplyDeleteIt not clear to me why CH=DM in your solution. Please explain.
Peter Tran
geting AH = a + d give us ang HBC = x
ReplyDeletedrawing DBM = x => ▲DMB = ▲HBC
(BM = BC = a, MCB = CMB, MBD = CBH = x : ASA )
In triangle MBC altitude is bisector, giving BM = a
To: c.t.o.e.
ReplyDeleteWhat a nice soluion, well done!
Ajit
To Joe
ReplyDeleteThanks
If BG is altitude of ABC than BG is median of MBC,DBH
=> MD = HC as diferences of equal parts
( I liked your solution of P 167 )
To c.t.e.o
ReplyDeleteYour solution is the most beautiful solution I've ever seen.
Trisect < CBD by BE and BF, E and F on AC and BE thus being an altitude as well
ReplyDeleteLet CE = e so that DE = EF = d-e and CF = 2e-d
< AFB = 90-x = < ABF so AB = AF = a+d = b - (2e-d) hence e = (b-a)/2......(1)
Now CB is tangential to Tr. ABE at B so,
a^2 = eb
Substituting for e in (1),
b(b-a)/2 = a^2 hence b^2 - ab - 2b^2 = 0
Solving the quadratic and rejecting the negative solution we have b = 2a hence 2x = 30 and x = 15
Sumith Peiris
Moratuwa
Sri Lanka
This is my solution, enjoy it:
ReplyDeletehttps://mega.nz/#!ZkwXWaRI!vn8pL5gHZ6hafXnKs0Y6TAgAyNAwUCcI-ECxJkwHtlI
Pedro Miranda