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

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## Wednesday, August 18, 2010

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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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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

Labels:
angle,
cevian,
right triangle,
sum of segments

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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