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Geometry ProblemClick the figure below to see the complete problem 505 about Right Triangle, Cevian, Sum of Segments, Angles.
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?Of course, this makes x=15 deg.Vihaan
draw DBM = x => BM = a, get AH = a + d , H on AC=> CH = DM => HM = d => MA = a => MBA = 2x, B = 6x=> x = 15°
To c.t.e.oIt not clear to me why CH=DM in your solution. Please explain.Peter Tran
geting AH = a + d give us ang HBC = xdrawing 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.What a nice soluion, well done!Ajit
To JoeThanksIf 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.oYour 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 wellLet 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 = ebSubstituting for e in (1),b(b-a)/2 = a^2 hence b^2 - ab - 2b^2 = 0Solving the quadratic and rejecting the negative solution we have b = 2a hence 2x = 30 and x = 15Sumith PeirisMoratuwaSri Lanka